2nd PUC

Students can Download Basic Maths Exercise 15.3 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Basic Maths Question Bank Chapter 15 Circles Ex 15.3

Part – A

2nd PUC Basic Maths Circles Ex 15.3 One Mark Questions and Answers

1. Show that the circle.

Question 1.
x² + y² – 8x + 2y + 16 = 0 touches x= axis.
Answer:
Given g = -4, f = 1, & C = 16

Question 2.
x² + y² – 4x + 4y + 4 = 0 touches x-axis.
Answer:
Here g = -2, C = 4
Here (-2)² = 4 ⇒ g² = C
∴ The circle touches x-axis.

Question 3.
x² + y² + 8x + 6y + 9 = 0 touches y = axis.
Answer:
Here f = 3 & C = 9
& f² = C ⇒ (3)² = 9
⇒ Circle touches y-axis.

Question 4.
x² + y² – x + 4y + 4 = 0 touches y-axis.
Answer:
Here f= 2 & C = 4
& f² = C ⇒ (2)² = 4
⇒ Circle0 touches y-axis

Question 5.
Show that the circle x² + y² + 4x – 3y + 4 = 0 touches x- axis.
Answer:
Here g = 2 & C = 4
g² = C ⇒ (2)² = 4
⇒ Circle touches x-axis.

Question 6.
Show that the circle x² + y² – 3x + 8y + 16 = 0 touches y – axis.
Answer:
Here f = 4, C = 16 & f² = C
⇒ (4)² = 16 .
∴ Circle touches y-axis.

Part – B

2nd PUC Basic Maths Circles Ex 15.3 Two Marks Questions and Answers

Question 1.
Find the length of the chord of the circle x² + y² – 6x + 4y + 5 = 0 intercepted by x-axis.
Answer:
Centre of given circle = (3,-2), c = 5
Length of chord intercepted by x – axis = 2\(\sqrt{g^{2}-c}\) = 2\(\sqrt{9-5}\) = 2\(\sqrt{4}\) = 2.2 = 4 units.

Question 2.
Find the length of the chord of the circle x² + y² – 6x + 15y – 16 = 0 intercepted by the y-axis.
Answer:
centre = \(\left(3, \frac{-15}{2}\right)\) & C = -16
Length of the chord intercepted by y – axis = 2\(\sqrt{f^{2}-c}\) = 2\(\sqrt{\left(\frac{15}{2}\right)^{2}+16}\)

Question 3.
Find the length of the chord of the circle x² + y² + 3x – 2 = 0 intercepted by y-axis.
Answer:
Centre = \(\left(-\frac{3}{2}, 0\right)\) c = -2
Length of the chord by y – axis = 2\(\sqrt{f^{2}-c}\) = 2\(\sqrt{0+2}\) = 2\(\sqrt{2}\) units.

Question 4.
Find the length of the chord of the circle x² + y² + 3x – y – 6 = 0 intercepted by the y-axis.
Answer:
Centre = \(\left(-\frac{3}{2}, \frac{1}{2}\right)\) c = -6
Length of the chord by axis = 2 \(\sqrt{f^{2}-c}\) = 2\(\sqrt{\frac{1}{4}+6}\) = 2\(\sqrt{\frac{25}{4}}\) = 5 units

Question 5.
Find the length of the chord of the circle x² + y² + 4x + 6y – 12 = 0 and x + 4y – 6 = 0.
Answer:

Question 6.
Find the length of the chord intercepted by the circle x² + y² – 8x – 6y – 6 = 0 and the line x – 7 y – 8 = 0.
Answer:

Question 7.
Find the length of the chord intercepted by the circle x² + y² – 6x – 2y + 5 = 0 and the line x – y + 1 = 0.
Answer:

Question 8.
Prove that the length of the chord x + 2y = 5 of the circle x² + y² = 9 is 4 units.
Answer:
Length of the chord intercepted by x – axis = 2\(\sqrt{g^{2}-c}\)
Here g = 5, f = -3, c = 9
∴ Length = \(2 \cdot \sqrt{25-9}=2 \sqrt{16}\)
= 2(4) = 8 units
Hence proved.
Centre of the given circle = (0,0), r = 3 units line = x + 2y – 5 = 0
\(d=\left|\frac{0+2(0)-5}{\sqrt{1^{2}+2^{2}}}\right|=\frac{5}{\sqrt{5}}=\sqrt{5}\)
∴ Length of the chord = \(2 \cdot \sqrt{r^{2}-d^{2}}\)
\(=2 \cdot \sqrt{(3)^{2}-(\sqrt{5})^{2}}=2 \sqrt{9-5}=2.2=4 \text { units. }\)
Hence proved.

Part – C

2nd PUC Basic Maths Circles Ex 15.3 Five Marks Questions and Answers

Question 1.
Show that the line 3x + 4y + 7 = 0 touches the circle x² + y² – 4x – 6y – 12 = 0 and find the point of contact.
Answer:
Given centre = (2, 3) and r = \(\sqrt{4+9+12}\) = \(\sqrt{25}\) = 5
If the length of the perpendicular is equal to radius of the circle than the line touches the circle

∴ The line 3x + 4y + 7 touches the circle
Any line perpendicular to 3x + 4y + 7 = 0 is of the form 4x – 3y + k = 0
It passes through (2,3) ∴ 8 – 9 + k = 0 ⇒ k = 1
∴ The line is 4x – 3y + 1 = 0
Point of contact is the intersection of 3x + 4y + 7 = 0 and 4x – 3y + 1 = 0

∴ The point of contact is (-1,-1)

Question 2.
Find k if the line 3x + y + k = 0 touches the circle x² + y² – 2x – 4y – 5 = 0.
Answer:
Given C = (1, 2) C = -5
Radius = Length of the perpendicular from (1, 2) to 3x + y + k = 0.

Question 3.
Find k if the 2x + y + k = 0 touches the circle x² + y² + 6x + 2y + 5 = 0.
Answer:
Hence centre = (-3,-1), c = 5
\(r=\sqrt{g^{2}+f^{2}+c}=\sqrt{9+1-5}=\sqrt{5}\)
r = length of the perpendicular from (-3,-1) to the line 2x + y + k = 0
\(r=\sqrt{g^{2}+f^{2}+c}=\sqrt{9+1-5}=\sqrt{5}\)

Question 4.
Find k if the line 4x – y + k = 0 touches the circle x² + y² + 4x – 8y + 3 = 0.
Answer:
Here centre = (-2,4), r = \(\sqrt{4+16-3}\) = \(\sqrt{17}\)
radius = length of perpendicular from (-2, 4) to the line 4x – y + k = 0
∴ \(\sqrt{17}\) = \(\left|\frac{4(-2)-4+k}{\sqrt{16+1}}\right|\)
-12 + k = ±\((\sqrt{17})^{2}\) ⇒ -12 + K = ±17 ⇒ k = 28 or -5

Question 5.
Find k such that the line x – 2y + k = 0 touches the circle x² + y² + 3x – 2y + 2 = 0.
Answer:
Here centre \(\left(-\frac{3}{2}, 1\right)\), \(r=\sqrt{\frac{9}{4}+1-2}=\sqrt{\frac{5}{4}}\)
r = length of perpendicular from \(\left(-\frac{3}{2},-1\right)\) to the line x – 2y + k = 0

Students can Download Basic Maths Exercise 15.2 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Basic Maths Question Bank Chapter 15 Circles Ex 15.2

Part – A

2nd PUC Basic Maths Circles Ex 15.2 One Mark Questions and Answers

Question 1.
Find the centre and the radius of the circle.
(a) x² + y² – 4x – y – 5 = 0
(b) 3x² + 3y² – 6x – 12y – 2 = 0
(c) (x – 2) (x – 4) + (y – 1) (y – 3) = 0
(d) x² + y² – 2x cosα – 2y sinα = 1
Answer:
(a) x² + y² – 4x – y – 5 = 0 Comparing with
x² + y² + 2gx + 2fy + C = 0 We get
2g = – 4, 2f = – 1 ⇒ g = -2, f = \(-\frac{1}{2}\)
∴ C=(-8,-1) = (2, \(\frac{1}{2}\))
\(r=\sqrt{g^{2}+f^{2}-C} ; \quad r=\sqrt{(-2)^{2}+\left(-\frac{1}{2}\right)^{2}-(-5)}=\sqrt{4+\frac{1}{4}+5}=\sqrt{\frac{16+1+20}{4}}=\frac{\sqrt{37}}{2} \text { units }\)

(b) Given 3x² + 3y² – 6x – 12y – 2 = 0, divide by 3
x² + y² – 2x – 4y – \(\frac{2}{3}\) = 0
Here g = -1, f = -2, & c = \(-\frac{2}{3}\)
∴Centre = (-g, -f) = (1, 2) & r =\(\sqrt{g^{2}+f^{2}-C}\)
\(=\sqrt{(-1)^{2}+(-2)^{2}-\left(-\frac{2}{3}\right)}\)
r = \(\sqrt{1-4-\frac{2}{3}}=\sqrt{\frac{17}{3}}\) units.

(c) Given (x – 2) (x – 4) + (y – 1) (y – 3) = 0
⇒ x² – 6x + 8 + y² – 4y + 3 = 0
⇒ x² + y² – 6x – 4y + 11 = 0
Here g = -3, f= -2, C = 11 & ∴ Centre = (3,2).
And r = \(\sqrt{9+4-11}=\sqrt{2}\) units

(d) Given x² + y² – 2x cos α – 2y sin α – 1 = 0
g = – cosα, f = – sinα, c = -1
∴ Centre = (cosα, sinα) & r = \(\sqrt{\cos ^{2} \alpha+\sin ^{2} \alpha+1}=\sqrt{1+1}=\sqrt{2}\)

Question 2.
If the radius of the circle x² + y² + 4x – 2y- k = 0 is 4 units find k.
Answer:
Given x² + y² + 4x – 2y – k = 0 & r = 4 units, K = ?
Here g = 2, f = -1 and C = – k
∴ Centre = (-2, 1) & r = \(\sqrt{g^{2}+f^{2}-c}\)
4 = \(\sqrt{(2)^{2}+(-1)^{2}-(-k)}\) ⇒ 4 = \(\sqrt{4+1+k}\) S.B.S
16 = 5 + k ⇒ k = 16 – 5 = 11

Question 3.
Find the other end of the diameter, if one of the diameter of the circle.
(a) x² + y² = 25 is (5,0)
(b) x² + y² + 4x – 6y – 12 = 0 is (-5, -1)
(c) x² + y² – 6x + 2y = 31 is (7,4)
Answer:
(a) Given x² + y² = 25, A = (5,0) Let (x₂ y₂) = ?
Here C =(0,0) & We know that centre is the midpoint of the diameter.

⇒ 5 + x₂ = 0 0 + y₂ = 0
⇒ x₂ = -5 y₂ = 0
∴ The other end of the diameter B = (-5,0).

(b) Given x² + y² + 4x – 6y – 12 = 0,

Here C = (-2,3)
∴ By mid point formula we have

⇒ x₂ = -4 + 5 = 1 & y₂ = 6 + 1 = 7
∴ The other end of the diameter B = (1,7)

(c) Given x² + y² – 6x + 2y – 31 = 0, A = (7,4), B = (x₂, y₂) = ?
Here C = (3,-1)

⇒ 7 + x₂ = 6 & 4 + y₂ = -2
⇒ x₂ = -1 y₂ = -6
∴ Other end of the diameter B = (-1,-6).

Question 4.
If(a, b) and (-5, 1) are the two end points of diameter of the circle x² + y² + 4x – 4y = 2. Find the value of a and b.
Answer:
Given (x₁ y₁) = (a, b) (x₂, y₂) = (-5, 1) & Center of the given circle = (-2,2).
Equation of the circle when ends of diameter is given is
(x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0.
(x – a) (x + 5) + (y – b) (y – 1) = 0.
⇒ x² – 5a – ax + 5x + y² – by – y + b = 0.
x² + y² + x (5 – a) – y(b + 1) + (b – 5a) = 0 comparing.
This by given circle x² + y³ + 4x – 4y – 2 = 0 we get
5 – a = 4, b + 1 = 4
⇒ 5 – 4 = a, b = 4 – 1
∴ a = 1 & b = 3.

Question 5.
If x² + y² – 4x – 8y + k = 0 represents a point circle find k
Answer:
Given x² + y² – 4x – 8y + k = 0 is a point circle
⇒ r = 0, g = -2, f = -4 & c = k
and \(\sqrt{g^{2}+f^{2}-C}=r\)
\(\sqrt{4+16-\mathrm{k}}=0 \quad=\sqrt{20-\mathrm{k}=0}\) ⇒ k = 20 S.B.S

Question 6.
If x² + y² + ax + 2y + 5 = 0 represent a point circle find a
Answer:
Given x² + y² + ax + 2y + 5 = 0 is a point circle.
⇒ r = 0, g = \(+\frac{a}{2}\) f = 1, C = 5
and \(\sqrt{g^{2}+f^{2}-C}=r=0\)

Question 7.
If x² + y² + ax + by = 3 represents a circle with centre at (1, -3), find a and b.
Answer:
Given x² + y² + ax + by – 3 = 0 & center = (1, -3), a = ?, b = ?
Here \(\frac { a }{ 2 }\) = 1 & \(\frac { b }{ 2 }\) = -3
⇒ a = 2 & b = – 6.

Question 8.
Find the unit circle concentric with the circle x² + y² – 8x + 4y = 8.
Answer:
Here unit circle ⇒ r= 1
& Center of concentric circle x² + y² – 8x + 4y – 8 = 0
C = (4, -2) & r =\(\sqrt{(-4)^{2}+(2)^{2}-(C)}=\sqrt{16+4-C}=1\)
⇒ 20 – C = 19
⇒ C = 19
∴ The Equation of the circle is x² + y² – 8x + 4y + 19 = 0.

Part-B

2nd PUC Basic Maths Circles Ex 15.2 Two Marks Question and Answers

Question 1.
Find the equation of the circle whose centre is same as the centre of the circle x² + y² + 6x + 2y + 1 = 0, and passing through the point (-2, -3).
Answer:
Given x² + y² + 6x + 2y + 1 = 0, P = (-2,-3)
Centre = C(-3,-1),
Let the equations of the required circle is x² + y² + 6x + 2y + C = 0
r = CP = \(\sqrt{(-2+3)^{2}+(-3+1)^{2}}=\sqrt{1^{2}+0}=1\)
∴ r = \(\sqrt{g^{2}+f^{2}-c}\)
l = \(\sqrt{9+1-c}\) ⇒ 1 = 10 -c ⇒ c = 9
∴ The required equation of the circle is
x² + y² + 6x + 2y + 9 = 0.

Question 2.
Find the equation of the circle whose centre is same as the centre of the circle x² + y² – 6x + 4y + 9 = 0 and Passsing through the point (-2,3)
Answer:
Given x² + y² – 6x + 4y + 9 = 0, & P = (-2,3)
Centre = c(3,-2), P = (-2,3)
r = cp = \(\sqrt{(-2-3)^{2}+(3-(-2))^{2}}=\sqrt{25+25}=\sqrt{50}\)
Let equation of the required circle is
x² + y² – 6x + 4y + c = 0
r = \(\sqrt{g^{2}+f^{2}-c}=\sqrt{9+4}\) ∴ \(\sqrt{50}=\sqrt{13-c}\) ⇒ 50 = 13 – c ⇒ c = – 37
∴ The required circle is x² + y² – 6x + 4y – 37 = 0.

Question 3.
Find the equation of the circle whose centre is (-2, 3) and passing through the centre of the circle x² + y² – 6x + 4y+9=0.
Answer:
Given c= (-2,3) & given x² + y² – 6x + 4y + 9 = 0. P = (3,-2)
r = cp = \(\sqrt{(3+2)^{2}+(-2-3)^{2}}=\sqrt{25+25}=\sqrt{50}\)
Also r = \(\sqrt{g^{2}+f^{2}-c}\)
\(\sqrt{50}=\sqrt{4+9-c}\) ⇒ 50 = 13 – C ⇒ c = -37
∴ Required circle with c = (-2-3) & r = \(\sqrt{50}\) is (x + 2)² + (y – 3)² = \((\sqrt{50})^{2}\)
x² + y² + 4x – 6y – 37 = 0.

Question 4.
Find the equation of the circle passing through the centre of the circle x² + y² – 2x – 4y – 20 = 0 and centre at (4, -2).
Answer:
Given x² + y² – 2x – 4y – 20 = 0 & c = (4, -2)
∴ P= (1, 2)
r = \(\sqrt{g^{2}+f^{2}-c}=\sqrt{16+4-(-c)}=\sqrt{20-c}\)
But r = CP = \(\sqrt{(4-1)^{2}+(-2-2)^{2}}=\sqrt{3^{2}+(-4)^{2}}=\sqrt{25}=5\)
∴ r = \(\sqrt{20-c}\) ⇒ 5 = \(\sqrt{20-c}\)
⇒ 25 = 20 – c ⇒ c = -5
∴ The required equation of the circle is x² + y² – 8x + 4y – 5 = 0.

Question 5.
Find the equation of the circle two of whose diameters are x + y = 3 and 2x + y = 2 and passing through the centre of the circle x² + y² – 4x + 2y – 1 = 0.
Answer:
Given x² + y² – 4x + 2y – 1 = 0
P = (2,-1)
& centre of the circle is point of intersection of diameters
x + y = 3 – (1) & 2x + y = 2 – (2)
eqn. (2) – eqn. (1) gives x = -1 & y = 3 – x = (3 – C – 1) = 4
∴ centre = c(-1,4)
r = CP = \(\sqrt{(-1-2)^{2}-\left(4-(-1)^{2}\right.}-\sqrt{(-3)^{2}+5^{2}}=\sqrt{9+25}=\sqrt{34}\)
∴ the eqn. of the circle with c = (-1,4) & r = \(\sqrt{34}\) is
(x + 1)² + (y – 4)² = \((\sqrt{34})^{2}\)
x² + 1 + 2x + y² + 16 – 8y = 34
x² + y² + 2x – 8y – 17 = 0.

Question 6.
Find the equation of the circle concentric with the centre of circle x² + y² – 2x + 2y – 1 = 0 and having double its area.
Answer:
Given x² + y² – 2x + 2y – 1 = 0
c = (1, -1) & r_ı = \(\sqrt{1+1-(-1)}=\sqrt{3}\)
Given A₂, = 2A,

r₂² =2 . \((\sqrt{3})^{2}\) = 2.3 = 6 ⇒ r₂ = \(\sqrt{6}\)
Equation of the required circle with c = (1,-1) & r = is (x – 1)² + (y + 1)² = \((\sqrt{6})^{2}\).
x² + 1 – 2x + y² + 1 + 2y = 6 ⇒ x² + y² – 2x + 2y – 4 = 0.

Question 7.
Find the equation of the circle concentric with the centre of the circle 3x² + 3y² – 6x + 9y – 2 = 0 and \(\frac { 2 }{ 3 }\) having of its area.
Answer:
Given 3x² + 3y² – 6x + 9y – 2 = 0, divide by 3
x² + y² – 2x + 3y – \(\frac { 2 }{ 3 }\) =0
centre = (1, \(\frac{-3}{2}\)) & r = \(\sqrt{1+\frac{9}{4}-\frac{2}{3}}=\sqrt{\frac{47}{12}}\)
Also given A₂ = \(\frac{2}{3}\) A₁
πr₂² = \(\frac{2}{3}\) . πr₂² ⇒ r₂² = \(\frac{2}{3} \times \frac{47}{12}=\frac{47}{18}\)
∴ Required circle with centre \(\left(1,-\frac{3}{2}\right), r_{2}-\sqrt{\frac{47}{18}} \text { is }(x-1)^{2}+\left(y+\frac{3}{2}\right)^{2}=(\sqrt{\frac{47}{18}})^{2}\)
x² + 1 – 2x + y² + 3y + \(\frac{9}{4}=\frac{47}{18}\)
x² + y² – 2x + 3y + \(\frac{23}{36}\) = 0 __________ × 36
36x² + 36y² – 72x + 108y + 23 = 0.

Question 8.
Find the equation of the diameter of the circle x² + y² + 6x – 2y = 6 which when produced passes through the point (1, -2).
Answer:
Given x² + y² + 6x – 2y – 6 = 0 & P (1, -2) centre = (-3, 1)
Slope of the diameter = m = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{1-(-2)}{-3-1}=\frac{3}{-4}\)
∴ Equation of the diameter with (1, -2) & m = \(\frac{-3}{4}\)
y – y₁ = m(x – x₁)
y + 2 = \(\frac{-3}{4}\)(x-1)
4y + 8 = -3x + 3 ⇒ 3x + 4y + 5 = 0.

Question 9.
Find the equation of the diameter of the circle 2x² + 2y² + 3x – 5y – 1 = 0 which when produced passes through the point (-1,2).
Answer:
Given 2x² + 2y² + 3x – 5y – 1 = 0 & p = (-1,2)
⇒ x² + y² + \(\frac { 3 }{ 2 }\)x – \(\frac { 5 }{ 2 }\)y – \(\frac { 1 }{ 2 }\) = 0
c = \(\left(-\frac{3}{4}, \frac{5}{4}\right)\)

Equation of the diameter with p(-1,2), & m = -3.
y – y₁, = m(x – x₁); y – 2 = -3(x + 1); y – 2 = -3x – 3; 3x + y + 1 = 0.

Question 10.
Show that the line 4x – y=17 passes through the centre of the circle x² + y² – 8x + 2y=0.
Answer:
Given x² + y² – 8x + 2y = 0 & line 4x – y = 17 c = (4, -1)
C must lie on the line 4x – y = 17
4(4)-(-1) = 17
16 + 1 = 17 ⇒ 17 = 17
Hence the line 6x – y = 17 passes through the centre.

Question 11.
Find the length of the chord of the circle x² + y² – 6x + 15y – 16 = 0 intercepted by the x-axis.
Answer:
Given x² + y² – 6x + 15y – 6 = 0
C = \(\left(3, \frac{-15}{2}\right)\) & c = -16
Length intercepted by x – axis = 2\(\sqrt{g^{2}-c}\)
\(=2 \sqrt{(3)^{2}-(-16)}=2 \sqrt{9+16}=2 \sqrt{25}=10\)
∴ Length = 10 units

Question 12.
Find the length of the chord of the circle x² + y² + 3x – y – 6 = 0 intercepted by the y-axis.
Answer:
Given x² + y² + 3x – y – 6 = 0
c = \(\left(-\frac{3}{2}, \frac{1}{2}\right)\) C=-6
Length intercepted by y-axis = 2\(\sqrt{f^{2}-c}\)

∴ Length = 10 units

Question 13.
Find the length of the chord of the circle x² + y² – 6x – 4y – 12 = 0 on the coordinate axes.
Answer:
Given x² + y² – 6x – 4y – 12 = 0
C = (3, 2) C = -12
Length intercepted by x-axis \(2 \sqrt{g^{2}-c}=2 \sqrt{9+12}=2 \sqrt{21}\) units
Length intercepted by y-axis = \(2 \sqrt{f^{2}-c}=2 \sqrt{4+12}=2 \sqrt{16}\) = 8units.

Part – C

2nd PUC Basic Maths Circles Ex 15.2 Five Marks Questions and Answers

Question 1.
Find the equation of the circle.
(a) Passing through the origin, having its centre on the x- axis and radius 2 units.
(b) Passing through (2, 3) having its centre on the x-axis and radius 5 units.
(c) Passing through the points (5, 1), (3, 4) and has its centre on the x – axis.
(d) Passing through the points (1, 2) and (2, 1) and has its centre on the y – axis.
(e) Passing through the points (0,5) and (6, 1) and has its centre on the line 12x + 5y = 25.
(f) Passing through the points (1, -4) and (5,2) and has its centre on the line x – 2y + 9 =0.
(g) Passing through the points (0, -3) and (0,5) and whose centre lies on x – 2y + 5 = 0.
(h) Passing through (1, 1) and (2, 2) and having radius 1.
Answer:
(a) Let A = (0,0), centre on x – axis ⇒ C = (-g, 0), r = 2 units
Required equation is x² + y² + 2gx + 2fy + c = 0
It passes their (0,0) ⇒ c = 0, r = \(\sqrt{g^{2}+f^{2}-c}\)
2 = \(\sqrt{g^{2}+0-0}\) ⇒ g = ±2
Equation of the circle is x² + y² ± 4x = 0.

(b) Let the general equation of the circle be x² + y² + 2gx + 2fy + c = 0.
It passes through (2,3) ⇒ (2)² + (3)² + 2g(2) + 2f(3) + c = 0
4g + 6f + c + 13 = 0 ….(1)
Centre (-g, -f) lies on x – axis ⇒ f = 0 …(2)
r = \(\sqrt{g^{2}+f^{2}-c}\) = 5 = \(\sqrt{g^{2}-c}\) = 25 = g² – C.
⇒ g² – 25 = c. ….(3)
Solving 1, 2 & 3 we get g² – 25 + 4g + 13 = 0
g² + 4g – 12 = 0
⇒ (g + 6) (g – 2) = 0 ⇒ g = -6 or 2
When g = -6, c = (-6)² – 25 = 36 – 25 = 11
g = 2, c = (2)² – 25 = 4 – 25 = -21 .
∴ The required circles are x² + y² – 12x + 11 = 0
x² + y² + 4x – 21 = 0.

(C) Let the required equation of the circle is x² + y² + 2gx + 2fy + c = 0
It passes through (5, 1) & (3, 4)
(5, 1) (5)² + (1)² + 2g(5) + 2f (1) +C = 0
10g + 2f + C + 26 = 0 …..(1)
(3,4) 3² + 4² + 2g (3) + 2f (4) +c= 0
68 + 8f + C + 25 = 0 …..(2)
& the centre (- g, – f) lies on x-axis ⇒ f = 0 ….(3)
Solving 1 & 2 we get

⇒ 2x² + 2y² – x – 47 = 0.

(d) Let required equation of the circle is x² + y² + 2gx + 2fy + c = 0
It passes through the points (1,2) & (2, 1) & centre lies on y-axis ⇒ g = 0 ….(1)
(1,2) 1² + 2² + 2g(1) + 2f(2) + c = 0 ⇒ 4f + C + 5 = 0 ….(2)
(2, 1) 2² + 1² + 2g(2) + 2f(1) + c = 0 ⇒ 2f+c + 5 = 0 ….(3)
Equation (3) – eqn. (2) gives 2f = 0 ⇒ f = 0
g = 0, f = 0 ⇒ c= -5
∴ Required equation of the circle is x² + y² – 5 = 0

(e) Let required equation of the circle is x² + y² + 2gx + 2fy + c = 0
It passes through (0,5) & (6, 1)
(0,5) 0 + 5² + 2g(0) + 2f(5) + c = 0
1f + C + 25 = 0 …(1)
(5, 1) 6² + 1² + 2g (6) + 2f (1) + c = 0
12g + 2f + c + 37 = 0 …..(2)
Centre (-g, -f) lies on the line 12x + 5y – 25 = 0
-12g – 5f – 25 = 0 ….(3)
Eqn. 2 – eqn. (1) gives
12g – 8f + 12 = 0 …..(4)
Adding 3 & 4 we get
12g – 13f – 13 = 0 ⇒ f = -1
From eqⁿ (1) ⇒ 10(-1) + C + 25 = 0 ⇒ C = – 15
∴ -12g = 5f + 25
-12g = 20 ⇒ g = \(\frac{20}{-12}=\frac{-5}{3}\)
∴ The required equation of the circle is
x² + y² + 2\(\left(-\frac{5}{3}\right)\)x + 2(-1)y – 15 = 0 ⇒ × 3
3x² + 3y² – 10x – 6y – 45 = 0

(f) Let the required equation of the circle is x² + y² + 2gx + 2fy + c = 0 but it passes through (1,-4) & (5,2) & centre (-g, -f) lies on the line
⇒ -g + 2f + 9 = 0 …… (1)
(1,-4) 1² + (-4)² + 2g(1) + 2f(-4) + c = 0
2g – 8f + C + 17 = 0 ….. (2)
⇒ (5,2) 5² + 2² + 2g(5) + 2f(2) +C = 0
10g + 4f + C+ 29 = 0 ….. (3)
Eqn. (3) – eqn. (2) gives 8g + 12f + 12 = () – 4
2g + 3f + 3 = 0 …..(4)

2f + 9 = g ⇒ g = -6 + 9 = 3
C = -2g + 8f – 17
C = -6 -24 – 17 = – 47
∴ The required equation of the circle is
x² + y² + 2 (3)x + 2 (-3)y – 47 = 0
x² + y² + 6x – 6y – 47 = 0

(g) Let the equation of required circle is x² + y² + 2gx + 2fy + c = 0 it passes
Through (0, -3) & (0,5) & centre (-g, -f)
Lies on the line x – 2y + 5 = 0 ⇒ -g + 2f + 5 = 0 …..(1)
(0,-3) 0 + 9 + (-6f) + c = 0
-6f + 9 + c = 0 …..(2)
(0,5) 0 + 25 + 0 + 10f + C = 0
10f + C + 25 = 0 …. (3)
∴ eqn. (3) – eqn. (2) gives 16f + 16 = 0 ⇒ f =-1
2f + 5 = g ⇒ g= -2 + 5 = 3
c = 6f – 9 = -6 -9 = -15 ⇒ c = -15
∴ The equation of the circle with g = 3, f = -1 & C = -15 is
x² + y² + 2 (3) x + 2 (-1)y – 15 = 0
x² + y² + 6x – 2y – 15 = 0

(h) Let the equation of the required circle is x² + y² + 2gx + 2fy + c = 0
If passes through (1, 1) & (2,2)
(1,1) 1² + 1² + 2g(1) + 2f(1) + c = 0
2g + 2f + C + 2 = 0 ….. (1)
(2, 2) 4 + 4 + 2g(2) + 2f (2) + c = 0
4g + 4f + c + 8 = 0 …… (2)
Also r = 1 ⇒ \(\sqrt{g^{2}+f^{2}-c}\) =1 ⇒ g² + f² – c = 1 …… (3)
Eqn. (2) – eqn. (1) gives 2g + 2f + 6 = 0
g + f + 3 = 0
g = (-3 -f) ….(4)
Adding 1 and 3 we get
g² + f² + 2g + 2f = -1
(-3 -f)² + f² + 2 (-3-f) + 2f + 1 = 0

2f² + 6f + 4 = 0 ÷ 2
f² + 3f + 2 = 0
(f + 2) (f + 1) = 0 ⇒ f = -1 or -2
When f = -2, g = -3-f = -3 + 2 = -1
When f = -1, g = -3-f = -3 + 1 = -2
C = g²+ f² -1 = 1 + 4 – 1 = 4
∴ Required circles are x² + y² – 2x – 4y + 4 = 0 and x² + y² – 4x – 2y + 4 = 0.

Question 2.
Find the equation of the circle passing through the points.
(a) (0, 2), (3, 0), (3, 2)
(b) (1, 1), (-2, 2), (-6,0)
(c) (1, 1), (5,-5), (6,-4)
(d) (1, 0), (3,0), (0, 2)
(e) (5, 7), (6, 6), (2,-2)
(0) (p, q), (p, 0), (0,)
(g) (0, 1), (2,3), (-2,5)
(h) (0,0), (a,0), (0, b)
Answer:
(a) Let the required equation of the circle is x² + y² + 2gx + 2fy + c = 0
This equation passes through (*0,2) (3,0) & (3,2)
(0,2) 0² + 2² + 2g(0) + 2 f(2) + c = 0
4f + 4 + c = 0 …. (1)
(3,0) 6g + 9 + c = 0 …. (2)
(3,2) 6g + 4f + 13 + c = 0 …. (3)
Solving (1), (2), (3) we get
(2) – (1) gives g – 4f + 5 = 0
(3) – (2) gives 4f + 4 = 0 f = -1
6g = 4(-1) -5 = -9 ⇒ g =\(-\frac{9}{6}=-\frac{3}{2}\).
C = -4f – 4 = 4 – 4 = 0
∴ Required equation of the circle is

(b) (1, 1) (-2,2) (-6,0)
The general equation of the circle x² + y² + 2gx + 2fy +C = 0
This equation passes through (1, 1) (-2,2) & (-6,0)
(1, 1) 1 + 1 + 2g + 2f + c = 0
2g + 2f + 2 + c = 0 …. (1)
(-2,2) 4 + 4 – 4g + 4f + c = 0
-4g + 4f + 8 + c ….. (2)
(-6,0) 36 + 0 – 12g + c = 0
-12g + 36 + c = 0 …. (3)
(2) – (1) gives -6g + 2f + 6 = 0 … (4) × 2

∴ The required eqn. is x² + y² + 4x + 6y – 12 = 0

(c) (1, 1) (5,-5) (6,-4)
The general equation of the circle
x² + y² + 2gx + 2fy + c = 0
This equation passes through (1, 1) (5,-5) (6, -4)
(1; 1) 2g + 2f + C + 2 = 0 ….. (1)
(5,-5) 10g – 10f + C + 50 = 0 …. (2)
(6,4) 12g + 8f + C + 52 = 0 …. (3)
(2)-(1) gives 8g – 12f + 48 = 0 ….. (4)
(3) – (2) gives 2g + 18f + 2 = 0 …. (5)
Solving (4) & (5) we get f = \(\frac { 10 }{ 11 }\) g = \(\frac { 111 }{ 21 }\) & C = \(-\frac { 284 }{ 21 }\)
∴ the required equation is
x² + y² + \(\frac { 222x }{ 21 }\) + \(\frac { 20y }{ 21 }\) – \(\frac { 284 }{ 21 }\) = 0 – × 21
21x² + 21y² + 222x + 20y – 284 = 0

(d) Let the general eqn, x² + y² + 2gx + 2fy + c = 0. The circle passes through the points
(1, 0) (3,0) & (0,2)
(1,0) 2g + C + 1 = 0 …. (1)
(0,2) 4f + C + 4 = 0 ….. (2)
(3,0) 6g + C + 9 = 0 …… (3)
Solving 1, 2, 3 we get g = -2, f = \(-\frac{7}{4}\) & c = 3

∴ 2x² + 2y² – 8x – 7y + 6 = 0.

(e) Let the general eqn. x² + y² + 2gx + 2fy + c = 0. The circle passes through the points
(5,7) (6, 6) (2,-2)
(5,7) 10g + 14f + 74 + c = 0 … (1)
(6,6) 12g + 12f + C + 72 = 0 …(2)
(2,-2) 4g – 4f + C + 8 = 0 …(3)
Solving 1, 2 and 3 we get g = -2, f = -3 & c= -12
∴ Required equation is x² + y² – 4x – 6y – 12 = 0.

(f) (p, q) (p, 0) (0,9)
Let the general eqn. x² + y² + 2gx + 2fy + c = 0.
The circle passes through the points
(p, q) (p, 0) (0,q)
(p, q) p² + q² + 2pq + 2qf + c = 0 … (1)
(p, 0) p² + 2pg + c = 0 …… (2)
(0,q) q² + 2qf + c= 0 … (3)
1 – 2 gives 2qf = -q² ⇒ f = \(-\frac{q}{2}\)
1 – 3 gives 2pg = -p² ⇒ g= \(-\frac{p}{2}\) & c = 0
∴ Required eqn. is x² + y² – px – qy = 0.

(g) Let the general eqn, x² + y² + 2gx + 2fy + c = 0. The circle passes through the points.
(0, 1) (2,3) (-2,5)
(0, 1) 2f + C = -1 ….. (1)
(2,3) 4g + 6 + c = -13 … (2)
(-2,5) -4g + 10f +c= -29 … (3)
Solving 1, 2, 3 we get g = \(\frac{1}{3}\) , f = \(-\frac{10}{3}\), c = \(\frac{17}{3}\)
∴ Required equation is x² + y² + \(\frac{2}{3}\)x – \(\frac{20}{3}\)y + \(\frac{17}{3}\) = 0
3x² + y² + 2x – 20y + 17 = 0

(h) Let the general equation x² + y² + 2gx + 2fy + c = 0. The circle passes through the points (0,0) (a, 0) (0, b)
(0,0) c = 0 …. (1)
(a,0) 2ag + c = -a² ⇒ g = \(-\frac{a}{2}\)
(0, b) 2bf + c = -b² ⇒ f = \(-\frac{b}{2}\)
∴ Required equation is x² + y² – ax – by = 0.

Question 3.
Find the equations of the circles whose radius is 5 and which passes through the points on X-axis at distances 3 from the origin.
Answer:
Given r = 5, passes through points on x-axis at a distance 3 from the origin
⇒ The points are (3,0) and (-3,0)
Let x² + y² + 2gx + 2fy + c = 0 is required equation
(3,0) 6g + C + 9 = 0 … (1)
(-3,0) -6g + C + 9 = 0 …. (2)
r = 5; 25 = g² + f² -C
Adding 1 and 2 we get c = -9 and g = 0 and f = ±4
∴ The required equation is x² + y² + 8y – 9 = 0

Question 4.
A circle has radius 3 units and its centre lies on the line y = x – 1. Find the equation of the circle if it passes through (7,3).
Answer:
Given r = 3 and centre (-g,-f) lies on y = x – 1
⇒ -f = -g -1
g – f + 1 = 0 …. (1)
Equation passes through (7,3) ⇒ 14g + 6f + C + 58 = 0 …. (2)
r = 3 ⇒ r² = g² + f² -c = 9 … (3)
Solving 1, 2, 3 we get g = -7 or g = -4, c = 76
f = -6 or g = -3, c = 16
∴ We get two equations of circles
x² + y² -14x – 12y + 76 = 0
x² + y² – 8x – 6y + 16 = 0.

Question 5.
Find the equation of the circle cuts intercepts of the length ‘a’ and ‘b’ on axes and passes through the origin
Answer:
Intercepts a and b on axes that implies circle passes through the points (a, 0) and (0, b) passing through origin
⇒ c = 0
(a, 0) a² + 2ga = 0 ⇒ g = \(-\frac{a}{2}\)
(0, -b) b² + 2fb = 0 ⇒ f = \(-\frac{b}{2}\)
∴ The required equation is x² + y² – ax – by = 0.

Question 6.
Find the equation of the circle passing through the origin and lengths ‘a’ and ‘b’ on axes and passing through the origin.
Answer:
Positive x-intercept a i.e., (a,0)
Negative y-intercept bi.e., (0, -b)
Let x² + y² + 2gx + 2fy + c = 0 passes through origin ⇒ c = 0 …. (1)
(a,0) ⇒ a² + 2ga = 0 = g ⇒ \(-\frac{a}{2}\) .. (2)
(0, -b) ⇒ b² – 2fb = 0 = f ⇒ \(-\frac{b}{2}\) … (3)
∴ The required equation is x² + y² – ax + by = 0.

Question 7.
Find the length of the chord intercepted by the
(a) circle x² + y² – 8x – 6y = 0 and the line x – 7y – 8 = 0
(b) circle x² + y² = 9 and the line x + 2y = 3
(c) circle x² + y² – 6x – 2y + 5 = 0 and the line x – y+ 1 = 0
Answer:
(a) Given centre (4,3)
r= \(=\sqrt{16+9}=\sqrt{25}\) = 5
P = length of the perpendicular from (4, 3) to the line x – 7y – 8 = 0

Length of the chord = 2 . \(\sqrt{r^{2}-p^{2}}\)
\(=2 \sqrt{5^{2}-\left(\frac{5}{\sqrt{2}}\right)^{2}}=2 \sqrt{25-\frac{25}{2}}=2 \sqrt{\frac{50-25}{2}}=2 \cdot \frac{5}{\sqrt{2}}\)
L = 5\(\sqrt{2}\) units

(b) Centre (0,0), r = 3, Line x + 2y – 3 = 0

Length of the chord = \(2 \cdot \sqrt{r^{2} \cdot d^{2}}=2 \sqrt{9-\frac{9}{5}}=2 \sqrt{\frac{36}{5}}=\frac{12}{\sqrt{5}}\) units

(c) Centre = (3, 1) and r = \(\sqrt{9+1-5}=\sqrt{5}\) , line x – y + 1 = 0

Length of the chord = \(2 \sqrt{r^{2}-d^{2}}=2 \sqrt{5-\frac{9}{2}}=2 \cdot \frac{1}{\sqrt{2}} 5 \text { units }\)

Question 8.
Find the points of intersection of the circle.
(a) x² + y² = 9 and the line x + 2y = 3.
(b) x² + y² – 6x – 2y + 5 = 0 and the line x – y + 1=0.
(c) x² + y² + 4x + 6y – 12 = 0 and the line x + 4y – 6 = 0
Also find the length of the chord.
Answer:
(a) Given x = 3 – 2y and x² + y² = 9
(3 – 2y)² + y² = 9
5y² – 12y = 0
y(5y – 12) = 0 ⇒ y = 0, y = 12
⇒ x = 3, x = \(-\frac{9}{5}\)
The points are (3, 0) and \(\left(-\frac{9}{5}, \frac{12}{5}\right)\)
∴ Length of the chord = \(\sqrt{\left(3+\frac{9}{5}\right)^{2}+\left(\frac{12}{5}\right)^{2}}=\frac{\sqrt{720}}{5}\)

(b) Given
x + 1 = y and x² + y² – 6x – 2y + 5 = 0
x² + (x + 1)² – 6x – 2 (x + 1) + 5 = 0
2x² – 6x + 4 = 0
x² – 3x + 2 = 0
(x – 2)(x – 1) = 0 = x= 2 or 1
When x = 2, y = 3 and when x = 1, y = 2
∴ The points are (2, 3) and (1, 2) and length of the chord
= \(\sqrt{(2-1)^{2}+(3-2)^{2}}=\sqrt{1+1}=\sqrt{2}\)

(c) Given x = 6 – 4y and x² + y² + 4x + 6y – 12 = 0
(6 – 4y)² + y² + 4 (6 – 4y) + 6y – 12 = 0
17y² – 58y + 48 = 0
17y² – 34y – 24y + 48 = 0
17y (y – 2) – 24 (y – 2) = 0
(y – 2) (17 y – 24) = 0 ⇒ y = 2 or
When y = 2, x = 6 – 4y = 6 – 8 = -2
When y = \(\frac { 24 }{ 17 }\) x = \(6-\frac{96}{17}=\frac{102-96}{17}=\frac{6}{17}\)
∴ The points are (-2,2) and \(\left(\frac{6}{17}, \frac{24}{17}\right)\)

Part-D

2nd PUC Basic Maths Circles Ex 15.2 Six Marks Questions and Answers

Question 1.
Show that the following points are concylie.
(a) (0, 0), (1, 1), (5, -5), (6,-4)
(b) (2,-4), (3, -1), (3,-3), (0,0)
(c) (1.0), (2, -7), (8, 1), (9,-6)
Answer:
(a) First find the equation of the circle passing through the points (0,0) (1, 1) (5,-5) (6, -4)
(0,0) ⇒ C = 0 …. (1)
(1, 1) ⇒ 2g + 2f + 2 = 0 …. (2)
(5,-5) ⇒ 10g -10f + 50 + c = 0 …. (3)

∴ The equation of the circle is x² + y² – 6x + 4y = 0
Substituting the fourth point (6,-4) we get
36 + 16 – 36 – 16 = 0 ⇒ 0 = 0
∴ The points are concyclic.

(b) Let us find the equation of the circle passing through (2,-4) (3,-1) and (3, -3) we get
(2,-4) 4g – 8f + C + 20 = 0 … (1)
(3,-1) 6g – 2f+ C + 10 = 0) … (2)
(3,-3) 6g – 6f+c= -18 … (3)
Solving the above equations we get g = -1, f = 2 and C = 0
∴ Required circle is x² + y² – 2x + 4y = 0
Substitute the fourth point (0,0), we get 0 + 0 + 0 + 0 = 0
∴ The four points are concyclic

(c) Let us find the equation of the circle passing through (1, 0) (2, -7) and (8, 1) we get
(1,0) 2g + C + 2 = 0 … (1)
(2,-7) 4g – 14f + C + 53 = 0 … (2)
(8,1) 16g + 2F + C + 25 = 0 … (3)
Solving above 3 equations we get g = \(-\frac { 738 }{ 50 }\),f = \(\frac { 147 }{ 50 }\) and c = \(\frac { 196 }{ 25 }\)
∴ Required circle is x² + y² + \(\frac{246}{50} x\) + \(\frac{147}{25} y\) + \(\frac{196}{25} \) = 0
25x² + 25y² + 246 x + 147 y + 196 = 0
Substitute the fourth point (9,-6) we get
25(9)² + 25(-6)² + 246 (9) + 147 (-6) + 196 = 0
2025 + 900 + 2214 – 882 + 196 0
The points are not concyclic

Students can Download Basic Maths Exercise 15.1 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Basic Maths Question Bank Chapter 15 Circles Ex 15.1

Part – A

2nd PUC Basic Maths Circles Ex 15.1 One Mark Questions and Answers

Question 1.
Find the equation of the circle with centre ‘C’ and radius ‘r’ in each of the following:
(a) c(-3, 2) and r= 5 units
(b) c(0, 0) and r = 4 units
(c) c(-1,-2) and diameter = 25 units
(d) c(1, 1) and r = \(\sqrt{2}\) units
(e) c(a cos θ, b sin θ) and r = a units.
Answer:
(a)

Equation of the circle is(x – x₁)² + (y – y₁)² = r²
(x-(-3))² + (y – 2)² = 5²
(x + 3)² + (y – 2)² = 5²
x² + y² + 6x – 4y – 12 = 0.

(b) Given C = (0,0) & r = 4 units
The equation of the circle is
(x – 0)² + (y – 0)² = 4²
⇒ x² + y² = 16

(c) C = (-1,-2 )& r = \(\frac{\text { diameter }}{2}=\frac{25}{2}\)
The equation of the circle is
(x + 1)² + (y + 2)² = \(\left(\frac{25}{2}\right)^{2}\)
x² + 2x + 1 + y² + 4y +4 = \(\frac { 625 }{ 4 }\)
⇒ 4x² + 4y² + 8x + 16y – 605 = 0.

(d) C = (1, 1) r = \(\sqrt{2}\) units
Equation of the circle is
(x – 1)² +(y – 1)² = \((\sqrt{2})^{2}\)
⇒ x² + 1 – 2x + y² + 1 – 2y = 2
⇒ x² + y² – 2x – 2y = 0

(e) C = (a cos θ, a sin θ) and r = a units
Equation of the circle is (x – a cos θ)²+ (y – a sin θ)² = a²
⇒ x² + a² cos² θ – 2a x cos θ + y² + a² sin² θ – 2a y sin θ = a²
⇒ x² + y² – 2a x cos θ – 2 ay sin θ + a² (cos² θ + sin² θ) = a²
⇒ x² + y² – 2a x cos θ – 2a y sin θ + a² – a²= θ
⇒ x² + y² – 2a x cos θ – 2a y sin θ = 0 (∵  cos² θ+ sin² θ = 1)

Question 2.
Find the centre of the circle, two of the diameters are
(a) x + y = 2, x – y = 0
(b) 2x – 3y = 1, 3x – 2y = 2
(c) y = 0 and y = x – 5
Answer:
(a) Given x + y = 2, x – y = 0 solving these two equations we get

(b) Given 2x – 3y = 1 → × 3
3x – 2y = 2 → × 2

(c) y = 0 ⇒ 0 = x – 5 ⇒ x = 5
∴ centre = (5,0).

Question 3.
Write the equation of the point circle with centre at
(a) (4, -5)
(b) (-3, 2)
(c) (1,0).
Answer:
(a) Point circle means circle whose radius = 0
Given C = (4,-5) & r = 0
The point circle & (x – 4)² + (y + 5)² = 0
⇒ x² + y² – 8x + 10y + 41 = 0

(b) Centre (-3,2)
The point circle is (x + 3)² + (y – 2)² = 0
⇒ x² + y² + 6x – 4y + 13 = 0

(C) Centre (1,0)
∴ Circle is (x – 1)² + (y – 0)² = 0
x² – 2x + 1 + y² = 0 ⇒ x² + y² – 2x + 1 = 0.

Part -B

2nd PUC Basic Maths Circles Ex 15.1 Two Marks Questions and Answers

Question 1.
Find the equation of the circle.
(a) Two of the diameters are x + y = 6 and x + 2y = 4 and its radius is 10 units.
(b) Two of the diameters x + y = 4 and x – y = 2 and passing through the point (2,-1).
(c) Two of the diameters are 2x – 3y = 5 and 3x – 4y = 7 are the diameters of a circle of area 154 Sq. cm.
Answer:
(a) Given r = 10, & Diameters are x + y = 6 & x + 2y = 4
Solving

∴ Centre = (8,-2) & r = 10 units.
∴ Equation of the circle is (x – 8)² + (y + 2)² = 10
⇒ x² + y² – 16x + 4y – 32 = 0.

(b) Point of inter section of two diameters x + y = 4 & x – y = 2 is the centre.

∴ Centre C = (3, 1). Also given P(2, – 1)
r = CP = \(\sqrt{(2-3)^{2}-(-1-1)^{2}}=\sqrt{(-1)^{2}+(-2)^{2}}=\sqrt{1+4}=\sqrt{5}\)
∴ The required equation of the circle with centre (3, 1) & radius = \(\sqrt{5}\) is
(x – 3)² – (y – 1)² = \((\sqrt{5})^{2}\)
x² + 9 – 6x + y² + 1 – 2y = 5
⇒ x² + y² – 6x – 2y + 5 = 0.

(c) Point of intersection of the two diameters is centre
2x – 3y = 5 → × 3
3x – 4y = 7 → × 2

∴ C = (1, – 1)
Also Given Area of the circle = 154 sq. cm
⇒ πr² = 154
\(r^{2}=\frac{154}{\pi}=\frac{154}{22} \times 7=49\)
⇒ r = 7
∴ Required Equation of the circle with centre (1, -1) & r = 7 units is
(x – 1)² + (y + 1)² = 49
x² + 1 – 2x + y² + 2y + 1 = 49
x² + y² – 2x + 2y – 47 = 0.

Question 2.
Find the equation of the circle with centre at (-2, 1) and passing through the origin.
Answer:
Given C = (-2, 1) & Let P = (0,0)
r = CP = \(\sqrt{(0-(-2))^{2}+(0-1)^{2}}=\sqrt{2^{2}+(-1)^{2}}=\sqrt{4+1}=\sqrt{5}\)
∴ Required equation of the circle with centre C = (-2, 1) & r = \(\sqrt{5}\) units is
(x + 2)² + (y – 1)² = \((\sqrt{5})^{2}\)
⇒ x² + y² + 4x – 2y = 0.

Question 3.
Find the equation of the circle with centre at (2, 1) and passing through (0, -1).
Answer:
Given Centre = C (2, 1) & Let P = (0, -1)
Here r = CP = \(\sqrt{(0-2)^{2}+(-1-1)^{2}}=\sqrt{4+4}=\sqrt{8}\) units
Equation of the circle with centre (2, 1) & r = \(\sqrt{8}\) is
(x – 2)² + (y – 1)² = \((\sqrt{8})^{2}\)
x² + 4x + y² + 1 – 2y = 8
x² + y² – 4x – 2y – 3 = 0.

Question 4.
Find the equation of the circle described on the line joining the points A and B as diameter where.
(a) A(-5, 1) and B(1,3)
(b) A(2,0) and B(0, 2)
(c) A(3,4) and B(1, -2)
Answer:
(a)

Equation of the circle is (x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0
i.e., (x + 5) (x – 1) + (y – 1) (y – 3) = 0
⇒ x² + 5x – x – 5 + y² – 3y – y + 3 = 0
⇒ x² + y² + 4x – 4y – 2 = 0.

(b)

Equation of the circle is (x – x₁)(x – x₂) + (y – y₁) (y – y₂) = 0
i.e., (x – 2)(x – 0) + (y – 0) (y – 2) = 0
x² – 2x + y² – 2y = 0
⇒ x² + y² – 2x – 2y = 0

(c) Given A(3,4) B(1, -2)
∴ Required equation is (x – x₁)(x – x₂) + (y – y₁) (y – y₂) = 0
i.e., (x – 3) (x – 1) + (y – 4) (y + 2) = 0
x² – x – 3x + 3 + y² + 2y – 4y – 8 = 0
⇒ x² + y² – 4x – 2y – 5 = 0.

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Karnataka 2nd PUC Maths Question Bank Chapter 7 Integrals Ex 7.11

2nd PUC Maths Integrals NCERT Text Book Questions and Answers Ex 7.11

By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

Question 1.
\(\int_{0}^{\frac{\pi}{2}} \cos ^{2} x d x\)
Answer:

Question 2.
\(\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x\)
Answer:

Question 3.
\(\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x d x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x} d x\)
Answer:

Question 4.
\(\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{5} x d x}{\sin ^{5} x+\cos ^{5} x}\)
Answer:

Question 5.
\(\int_{-5}^{5}|x+2| d x\)
Answer:

Question 6.
\(\int_{2}^{8}|x-5| d x\)
Answer:

Question 7.
\(\int_{0}^{1} x(1-x)^{n} d x\)
Answer:

Question 8.
\(\int_{0}^{\frac{\pi}{4}} \log (1+\tan x) d x\)
Answer:

Question 9.
\(\int_{0}^{2} x \sqrt{2-x} d x\)
Answer:

Question 10.
\(\int_{0}^{\frac{\pi}{2}}(2 \log \sin x-\log \sin 2 x) d x\)
Answer:

Question 11.
\(\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin ^{2} x d x\)
Answer:

Question 12.
\(\int_{0}^{\pi} \frac{\mathbf{x} \mathbf{d} \mathbf{x}}{1+\sin \mathbf{x}}\)
Answer:

Question 13.
\(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^{7} x d x\)
Answer:

Question 14.
\(\int_{0}^{2 \pi} \cos ^{5} x d x=I\)
Answer:

Question 15.
\(\int_{0}^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1+\sin x \cos x} d x\)
Answer:

Question 16.
\(\int_{0}^{\pi} \log (1+\cos x) d x\)
Answer:

Question 17.
\(\int_{0}^{\mathrm{a}} \frac{\sqrt{\mathbf{x}}}{\sqrt{\mathbf{x}}+\sqrt{\mathbf{a}-\mathbf{x}}} \mathbf{d} \mathbf{x}\)
Answer:

Question 18.
\(\int_{0}^{4}|x-1| d x\)
Answer:

Question 19.
Show that \(\int_{0}^{a} f(x) g(x) d x=2 \int_{0}^{a} f(x) d x\) if f and g are defined as f(x) = f(a – x) and g(x) + g(a – x) = 4 . Choose the correct answer in Exercises 20 and 21
Answer:

Question 20.
The value \(\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\left(x^{3}+x \cos x+\tan ^{3} x+1\right) d x\) dx is……..
(A) 0
(B) 2
(C) π
(D) 1
Answer:

Question 21.
The Value of \(\int_{0}^{\frac{\pi}{2}} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x\)
(A) 2
(B) \(\frac{3}{4}\)
(C) 0
(D) -2
Answer:

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Karnataka 2nd PUC Maths Question Bank Chapter 7 Integrals Ex 7.10

2nd PUC Maths Integrals NCERT Text Book Questions and Answers Ex 7.10

Question 1.
\(\int_{0}^{1} \frac{x}{x^{2}+1} d x\)
Answer:

Question 2.
\(\int_{0}^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos ^{3} \phi d \phi\)
Answer:

Question 3.
\(\int_{0}^{1} \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x\)
Answer:

Question 4.
\(\int_{0}^{2} x \sqrt{x+2}\left(\text { put } x+2=t^{2}\right)\)
Answer:

Question 5.
\(\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^{2} x} d x\)
Answer:

Question 6.
\(\int_{0}^{2} \frac{d x}{x+4-x^{2}}\)
Answer:

Question 7.
\(\int_{-1}^{1} \frac{d x}{x^{2}+2 x+5}\)
Answer:

Question 8.
\(\int_{1}^{2}\left(\frac{1}{x}-\frac{1}{2 x^{2}}\right) e^{2 x} d x\)
Answer:

choose the correct answer in exercise 9 and 10

Question 9.
The value of the integral \(\int_{\frac{1}{3}}^{1} \frac{\left(x-x^{3}\right)^{\frac{1}{3}}}{x^{4}} d x is \)
(A) 6
(B) 0
(C) 3
(D) 4
Answer:

Question 10.
\(\text { If } f(x)=\int_{0}^{x} t \sin t d t, \text { then } f^{\prime}(x) \text { is }\)
(A) cos x +x sin x
(B) x sin x
(C) x cos x
(D) sin x + x cos x
Answer:

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Karnataka 2nd PUC Basic Maths Question Bank Chapter 16 Parabola Ex 16.1

Part – A

2nd PUC Basic Maths Parabola Ex 16.1 One or Two Marks Questions and Answers

Question 1.
Write the characteristics of the following parabolas.
(a) y² = 16x
(b) y² = -8x
(c) 3x² = -8y
(d) x² = 8y
(e) y² = 8x
(f) y² + 4x = 0
(g) 3x² + 4y = 0
(h) x² + 4x = 0
Answger:
(a) Compare y² = 16x with y² = 4ax

We get 4a = 16 ⇒ a = 4
The curve turns right side with vertex (0,0)
Focus S(a,0) = (4,0)
Directrix x = -a or x = -4 or x + 4 = 0
Axis x-axis (equation is y = 0)
Tangent y-axis (equation is x = 0)
Equation of LR is x = a ⇒ x = 4 or x – 4 = 0
Length of LR 4a = 16
Ends of LR (a, 2a) (a, -2a) = (4,8) and (4,-8)

(b) y² = -8x compare with y² = -4ax

The curve turns left hand side and 4a = 8 ⇒ a = 2
Vertex = (0,0)
Axis x-axis (equation is y = 0)
Tangent y-axis (equation is x = 0)
Focus S(-2,0)
Directrix, x = 2 or x – 2 = 0
Equation of LR x = -2 or x + 2 = 0
Length of LR = 4a = 8
Ends of LR (-a, 2a) (-a, -2a) = (-2, 4) (-2,-4)

(C) 3x² = -8y ⇒ x² = \(\frac {-8 }{ 3 }\) y compare this with

x² = -4ay ⇒ 4a = \(\frac { 8 }{ 3 }\) ⇒ a = \(\frac { 2 }{ 3 }\)
The curve turns downwards
Vertex V = (0,0)
Axis y-axis (equation is x = 0)
Focus S = \(\left(0,-\frac{2}{3}\right)\)
Tangent x-axis (equation is y = 0)
Directrix y = \(\frac { 2 }{ 3 }\) = or 3y – 2 = 0
Equation of LR y = –\(\frac { 2 }{ 3 }\) or 3y + 2 = 0
Length of LR = 42 = \(\frac { 8 }{ 3 }\) ; Ends of LR \(\left(\frac{4}{3},-\frac{2}{3}\right)\left(-\frac{4}{3},-\frac{2}{3}\right)\)

(d) x² = 8y compare this with x² = 4ay the curve turns upwards

4a = 8 ⇒ a = 2
Vertex, V = (0,0)
Axis y-axis (equation is x = 0)
Tangent x-axis (equation is y = 0)
Directrix y = -2 or y + 2 = 0
Equation of LR y = 2 or y – 2 = 0
Length of LR 4a = 8
Ends of LR (4, 2) (-4, 2)

(e) y² = 8x compare with y² = 4ax. The curve turns right side

4a = 8 ⇒ a = 2
Vertex, V = (0,0)
Axis, x-axis (equation is y = 0)
Tangent, y-axis (equation is x = 0)
Focus S = (2,0)
Directrix x = -2 or x + 2 = 0
Equation of LR x = 2
Length of LR 4a = 8
Ends of LR (2, 4) (2,-4)

(f) Compare y² = -4x with y² = -4ax. The curve turns

Left hand side
4a = 4 ⇒ a = 1
Vertex, V = (0,0)
Axis x-axis (equation is y = 0)
Tangent y-axis (equation is x = 0)
Directrix, x = 1 or x = -1 = 0
Equation of LR x = -1 or x + 1 = 0
Ends of LR (-1,2) (-1, -2)
Length of LR = 4a = 4

(g) 3x² + 4y = 0

x² = \(\frac{-4}{3} y\) compare with x² = – 4ay.
The curve turns downwards and 4a = \(\frac { 4 }{ 3 }\) ⇒ a = \(\frac { 1 }{ 3 }\)
Vertex, V = (0,0)
Axis y-axis (equation is x = 0)
Tangent x-axis (equation is y = 0)
Focus, s=(0,- \(\frac { 1 }{ 3 }\)) Directrix y = \(\frac { 1 }{ 3 }\) or 3y – 1=0
Equation of LR y = \(\frac { -1 }{ 3 }\) or 3y +1=0 Length of LR = 4a = \(\frac { 4 }{ 3 }\)
Ends of LR \(\left(-\frac{2}{3},-\frac{1}{3}\right)\left(\frac{2}{3},-\frac{1}{3}\right)\)

(h) x² + 16y = 0 compare with x² = -4ay

x² = -16y. The curve turns downward
4a = 16 ⇒ a = 4
Vertex, V = (0,0)
Axis y-axis (Equation is x = 0)
Tangent x-axis (Equation is y = 0)
Focus, S = (0, -4)
Directrix y = 4 or y – 4 = 0
Equation of LR y = -4 or y + 4 = 0
Length of LR = 4a = 16
Ends of LR (-8,-4) (8,-4).

Question 2.
If the length of the latus rectum of y² = 8kx is 4 find k
Answer:
Compare y² = 8kx with y² = 4ax
We get 4a = 8k ⇒ a = 2k
Given length of LR = 4
4a = 4 ⇒ 8k = 4 ⇒ k = \(\frac { 1 }{ 2 }\)

Question 3.
If the length of the latus rectum of x² = 4ky is 8, find the value of k.
Answer:
Compare x² = 4ky with x² = 4ay.
We get 4k = 4a ⇒ a=k
Given length of LR = 8
4a = 8
4k = 8
k = 2.

Part-B

2nd PUC Basic Maths Parabola Ex 16.1 Two or Three Marks Question and Answers

Question 1.
Find the equation of the parabola given that its.
(a) Vertex is (0,0) and focus is (4,0)
(b) Vertex is (0,0) and directrix is y = – 3
(c) Focus is (1,0) and directrix is x = – 1,
(d) Focus is (-4, 0) and directrix is x = 4
(e) Focus is (0, -3) and directrix is y=3
(f) Focus is (0, 6) and vertex is (0,0)
(g) Vertex is (0,0), axis is y axis and passes through \(\left(\frac{1}{2}, 2\right)\)
(h) Vertex is (0,0) axis is y axis and passes through (-1, -3)
Answer:
(a) Given V = (0,0) and Focus, S = (4,0)
It is a right handed parabola with
a = 4 and standard form is
y² = 4ax
Put a = 4, then
y² = 16x.

(b) Given V = (0,0) and directrix = y = -3
Focus = (0,3) ⇒ a = 3
It is a upward parabola and
standard form is x² = 4ay
Put a = 3, x² = 4.3y
⇒ x² = 12y

(C) Given Focus = S = (1,0) and directrix is
x =-15 ⇒ a = 1
it is a right handed parabola
and standard form is y² = 4ax put a = 1
⇒ y² = 4x

(d) Give Focus = S = (-4, 0) and directrix
is x = 4; S = (-4,0) ⇒ a = 4
It is a left handed parabola standard
form is y² = -4ax put a = 4
we get y² = -16x

(e) Given focus is (0, -3) and directrix is y = 3
⇒ a = 3 and curve turns
downwards and standard form is
x² = -4ay put a = 3,
we get x² = -12y

(f) Given focus S = (0,6) and Vertex, V = (0,0)
⇒ a = 6 and it is a upward parabola and
the standard form is x² = 4ay
Put a = 6
We get x² = 24y

(g) Given vertex V = (0,0) and axis is y-axis and it passes through \(\left(\frac{1}{2}, 2\right)\)
which is in I quadrant
∴ it is a upward parabola
x² = 4ay put x = \(\frac { 1 }{ 2 }\) and y = 2; \(\frac { 1 }{ 4 }\) = 4a 2 ⇒ \(\frac { 1 }{ 4 }\) = 8a ⇒ a = \(\frac { 1 }{ 32 }\)
∴ The required equation is x² = 4ay
x² =4. \(\frac{1}{32} y\) ⇒ x² = \(\frac{1}{8} y\) or 8x² – y = 0

(h) Vertex, V(0, 0) and y-axis and it passes through the point (-1, -3)
which is in third quadrant
∴ It is a downward parabola,
i.e., x² = -4ay
put x=-1 and y = -3; (-1)² = -4a(-3)
1 = 12a ⇒ a = \(\frac { 1 }{ 12 }\)
∴ Required equation is x² = -4ay
x² = -4.\(\frac { 1 }{ 12 }y\) ; 3x² + y = 0.

Part – C

2nd PUC Basic Maths Parabola Ex 16.1 Five Mark Questions and Answers

Question 1.
Define parabola and with usual notation Prove that y² = 4ax geometrically
Sol.
Definition of Parabola and other forms of parabola
Parabola is a conic section whose eccentricity is 1. That is, Parabola is the locus of the point which moves such that its distance from the fixed point (called the focus) is equal to its distance from the fixed line (called the directrix) We shall derive the equation of the parabola in its standard form.

Theorem: The equation of a Parabola with proper choice of co-ordinates axes is y² = 4ax
Proof: Let S be the focus and the line l be the directrix. Draw SZ through S and perpendicular to the directrix. Let O be the midpoint of the line segment SZ.

Let SZ = 2a (a > 0) ⇒ OZ = OS = a
We shall choose 0 as the origin and the line ZOS as x axis.
The line YOY’ through O, perpendicular to the x-axis will be y-axis.
With this choice of co-ordinate axes we have S(a, o) and Z(-a, 0)
The equation of the directrix is l is x = -a
Let P(x,y) be any point on the Parabola. Then by the definition of the Distance of P
from S = Distance of P from the line ‘l’.

|SP| = |PM|
SP² = ZN² since PM = ZN
SP² = (OZ + ON)²
(x – a)² + y² = (a + x)² (|OZ| = a, |ON| = x)
x² – 2ax + a² + y² = a² + 2ax + x²
y² = 4ax which is the equation of the parabola.

Note:
The equation y² = 4ax is also called standard form of the equation of the parabola.
Shape of the Parabola y² = 4ax
We shall note few observations from the equation y² = 4ax, which will help us to trace the curve parabola.
1. Ify is replaced by -y in the equation remains same, i.e., (-y)² = 4ax → y² = 4ax. This shows that if (x, y) is any point on the curve y² = 4ax, then (x, -y) is also a point on the curve. Thus, the curve is symmetric about the x – axis, i.e., the shape of the curve above the x axis is the mirror image (about the x axis) of the shape of the curve below the x – axis.

2. If x < 0, then y will be a negative quantity (note that a > 0) and therefore y² = 4ax will have no real solution for y. This shows that no part of the curve lies to the left side of the x-axis.

3. If y =0, the only value of x we get is zero. Thus the curve cuts the x axis at the origin(0,0)

4. If x = 0, we get y² = 0, which gives y = 0. Thus the curve cuts the y-axis at the origin and further the y-axis meets the curve only at the origin. That is, y – axis is a tangent to the curve at the origin.

5. For any point P(x, y) on the parabola we have
y² = 4ax ⇒ y = ±2\(\sqrt{\mathrm{ax}}\)
y = 2\(\sqrt{\mathrm{ax}}\) and y = -2\(\sqrt{\mathrm{ax}}\)
This shows that, as x increases from 0 to ∞, y also increases from 0 to ∞ or y decreases from 0 to -∞. Thus, the two branches of the parabola, laying on opposite sides of the x – axis, will extend to infinity towards the positive directions of the x – axis.

From the above discussion and by plotting few points, whose coordinates satisfy y² = 4ax, it is found the shape of the parabola is as shown in the following figure.

The origin 0 is called the Vertex of the parabola, y² = 4ax, it is also denoted by V(0,0).
The line ZSX is called the axis of the parabola and its equation is y = 0
The focus S (a,0) and the equation of the directrix is x – a i.e., X + a = 0 They – axis is called the tangent at the vertex – its equation is x = 0.

Note:

The distance between the vertex and the focus is the distance between the vertex and the directrix is equal to a and the distance between the directrix and the focus is 2 a.

Definition:
The chord passing through the focus and perpendicular to the axis of the parabola is called the latus rectum of the parabola.

In the figure LSL¹ is the latus rectum.
The points L and L¹ on the parabola are called end points of the latus rectum, the length is called the length of the latus rectum.
Clearly, the x – coordinates of L and L’ is a because OS = a. To find the corresponding coordinates, we shall put x = a in y² = 4ax, we get y = ±2a. Thus, y – coordinates of L and L¹ respectively.
L = (a, 2a) and L¹ = (a, -2a)
∴ consider |LL’| = \(\sqrt{(a-a)^{2}+(2 a+2 a)^{2}}=\sqrt{(4 a)^{2}}=4 a\)
Thus, the length of the latus rectum, LL¹ = 4a.

Note :
Any chord passing through the focus is called a focal chord. Focal chord need not be perpendicular to the axis of the parabola

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Karnataka 2nd PUC Maths Question Bank Chapter 7 Integrals Ex 7.9

2nd PUC Maths Integrals NCERT Text Book Questions and Answers Ex 7.9

Question 1.
\(\int_{-1}^{1}(x+1) d x\)
Answer:

Question 2.
\(\int_{2}^{3} \frac{1}{x} d x\)
Answer:

Question 3.
\(\int_{1}^{2}\left(4 x^{3}-5 x^{2}+6 x+9\right) d x\)
Answer:

Question 4.
\(\int_{0}^{\frac{\pi}{4}} \sin 2 x d x\)
Answer:

Question 5.
\(\int_{0}^{\frac{\pi}{2}} \cos 2 x d x\)
Answer:

Question 6.
\(\int_{4}^{5} e^{x} d x\)
Answer:

Question 7.
\(\int_{0}^{\frac{\pi}{4}} \tan x d x\)
Answer:

Question 8.
\(\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \csc x d x\)
Answer:

Question 9.
\(\int_{0}^{1} \frac{d x}{\sqrt{1-x^{2}}}\)
Answer:

Question 10.
\(\int_{0}^{1} \frac{d x}{1+x^{2}}\)
Answer:

Question 11.
\(\int_{2}^{3} \frac{d x}{x^{2}-1}\)
Answer:

Question 12.
\(\int_{0}^{\frac{\pi}{2}} \cos ^{2} x d x\)
Answer:

Question 13.
\(\int_{2}^{3} \frac{x d x}{x^{2}+1}\)
Answer:

Question 14.
\(\int_{0}^{1} \frac{2 x+3}{5 x^{2}+1} d x\)
Answer:

Question 15.
\(\int_{0}^{1} x e^{x^{2}} d x\)
Answer:

Question 16.
\(\int_{1}^{2} \frac{5 x^{2}}{x^{2}+4 x+3}\)
Answer:

Question 17.
\(\int_{0}^{\frac{\pi}{4}}\left(2 \sec ^{2} x+x^{3}+2\right) d x\)
Answer:

Question 18.
\(\int_{0}^{\pi}\left(\sin ^{2} \frac{x}{2}-\cos ^{2} \frac{x}{2}\right) d x\)
Answer:

Question 19.
\(\int_{0}^{2} \frac{6 x+3}{x^{2}+4} d x\)
Answer:


Question 20.
\(\int_{0}^{1}\left(x e^{x}+\sin \frac{\pi x}{4}\right) d x\)
Answer:

Choose the correct answer in Exercise 21 and 22

Question 21.
\(\int_{1}^{\sqrt{3}} \frac{d x}{1+x^{2}} \text { equals }\)
(A) \(\frac{\pi}{3}\)
(B) \(\frac{2 \pi}{3}\)
(C) \(\frac{\pi}{6}\)
(D) \(\frac{\pi}{12}\)
Answer:

Question 22.
\(\int_{0}^{\frac{2}{3}} \frac{d x}{4+9 x^{2}} \text { equals }\)
(A) \(\frac{\pi}{6}\)
(B) \(\frac{\pi}{12}\)
(C) \(\frac{\pi}{24}\)
(D) \(\frac{\pi}{4}\)
Answer:

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Karnataka 2nd PUC Maths Question Bank Chapter 7 Integrals Ex 7.8

2nd PUC Maths Integrals NCERT Text Book Questions and Answers Ex 7.8

Evaluate the following definite integrals as limit of sums

Question 1.
\(\int_{a}^{b} x d x\)
Answer:

Question 2.
\(\int_{0}^{5}(x+1) d x\)
Answer:

Question 3.
\(\int_{2}^{3} x^{2} d x\)
Answer:

Question 4.
\(\int_{1}^{4}\left(x^{2}-x\right) d x\)
Answer:

Question 5.
\(\int_{-1}^{1} e^{x} d x\)
Answer:

Question 6.
\(\int_{0}^{4}\left(x+e^{2 x}\right) d x\)
Answer:

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Karnataka 2nd PUC Maths Question Bank Chapter 7 Integrals Ex 7.7

2nd PUC Maths Integrals NCERT Text Book Questions and Answers Ex 7.7

Question 1.
\(\sqrt{4-x^{2}}\)
Answer:

Question 2.
\(\sqrt{1-4 x^{2}}\)
Answer:

Question 3.
\(\sqrt{x^{2}+4 x+6}\)
Answer:

Question 4.
\(\sqrt{x^{2}+4 x+1}\)
Answer:

Question 5.
\(\sqrt{1-4 x-x^{2}}\)
Answer:

Question 6.
\(\sqrt{x^{2}+4 x-5}\)
Answer:

Question 7.
\(\sqrt{1+3 x-x^{2}}\)
Answer:

Question 8.
\(\sqrt{x^{2}+3 x}\)
Answer:

Question 9.
\(\sqrt{1+\frac{x^{2}}{9}}\)
Answer:

Choose the correct answer in Exercises 10 to 11.

Question 10.
\(\sqrt{1+x^{2}} \) is equal to

Answer:

Question 11.
\(\int \sqrt{x^{2}-8 x+7} d x\) is equal to

Answer:

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Karnataka 2nd PUC Maths Question Bank Chapter 7 Integrals Ex 7.6

2nd PUC Maths Integrals NCERT Text Book Questions and Answers Ex 7.6

Integrate the functions in Exercises 1 to 22.

Question 1.
x sin x
Answer:
∫ x sin x . dx
= x(-cosx) +∫ 1 x cos x dx
= – x cos x + sin x + C

Question 2.
x sin 3x
Answer:

Question 3.
x²e^x
Answer:
∫ x²e^x.dx
= x²e^x -∫2x x e^x dx = x²e^x -2(xe^x -1 x e^x)
= x²e^x – 2x e^x + 2e^x +C
= e^x (x² – 2x + 2) + C

Question 4.
x log x
Answer:

Question 5.
x log 2x
Answer:

Question 6.
x² log x
Answer:

Question 7.
x sin^-1 x
Answer:

Question 8.
x tan^-1 x
Answer:

Question 9.
x cos^-1 x
Answer:

Question 10.
(sin^-1 x)²
Answer:

Question 11.
\(\frac{x \cos ^{-1} x}{\sqrt{1-x^{2}}}\)
Answer:

Question 12.
x sec^-2 x
Answer:
sec²x
= x tan x -∫tan x + dx
= x tan x + log |cos x| +C

Question 13.
tan^-1 x
Answer:

Question 14.
x(log x)²
Answer:

Question 15.
(x² + 1) log x
Answer:

Question 16.
e^x (sin x + cos x)
Answer:
∫e^x (sin x + cos x) dx
put f(x) = sin x
f’(x) = cos x
= ∫e^x (f(x) + f’(x)) dx = e’f(x) + C
=∫e^x (sinx + cosx) dx = e^x sin x + C

Question 17.
\(\frac{\mathbf{x} \mathbf{e}^{\mathbf{x}}}{(\mathbf{1}+\mathbf{x})^{2}}\)
Answer:

Question 18.
\(\mathbf{e}^{x}\left(\frac{1+\sin x}{1+\cos x}\right)\)
Answer:

Question 19.
\(e^{x}\left(\frac{1}{x}-\frac{1}{x^{2}}\right)\)
Answer:

Question 20.
\(\frac{(x-3) e^{x}}{(x-1)^{3}}\)
Answer:

Question 21.
e^2x sin x
Answer:

Question 22.
\(\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\)
Answer:

Question 23.
\(\int x^{2} e^{x^{2}} d x \text { equals }\)

Answer:

Question 24.
∫e^x sec x (1+ tan x) dx equals
(A) e^x cos x+ C
(B) e^x sec x+ C
(C) e^x sin x+ C
(D) e^x tan x+ C
Answer:
∫e^x sec x (1 + tan x) dx
= ∫e^x (sec x + sec x tan x) dx
put f (x 0 = sec x, f’ (x) = sec x tan x
= ∫ e^x (sec x + sec x tan x) dx = e^x sec x + C
Answer (B)