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## 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) x^{2} + y^{2} – 4x – y – 5 = 0

(b) 3x^{2} + 3y^{2} – 6x – 12y – 2 = 0

(c) (x – 2) (x – 4) + (y – 1) (y – 3) = 0

(d) x^{2} + y^{2} – 2x cosα – 2y sinα = 1

Answer:

(a) x^{2} + y^{2} – 4x – y – 5 = 0 Comparing with

x^{2} + y^{2} + 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 3x^{2} + 3y^{2} – 6x – 12y – 2 = 0, divide by 3

x^{2} + y^{2} – 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

⇒ x^{2} – 6x + 8 + y^{2} – 4y + 3 = 0

⇒ x^{2} + y^{2} – 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 x^{2} + y^{2} – 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 x^{2} + y^{2} + 4x – 2y- k = 0 is 4 units find k.

Answer:

Given x^{2} + y^{2} + 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) x^{2} + y^{2} = 25 is (5,0)

(b) x^{2} + y^{2} + 4x – 6y – 12 = 0 is (-5, -1)

(c) x^{2} + y^{2} – 6x + 2y = 31 is (7,4)

Answer:

(a) Given x^{2} + y^{2} = 25, A = (5,0) Let (x_{2} y_{2}) = ?

Here C =(0,0) & We know that centre is the midpoint of the diameter.

⇒ 5 + x_{2} = 0 0 + y_{2} = 0

⇒ x_{2} = -5 y_{2} = 0

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

(b) Given x^{2} + y^{2} + 4x – 6y – 12 = 0,

Here C = (-2,3)

∴ By mid point formula we have

⇒ x_{2} = -4 + 5 = 1 & y_{2} = 6 + 1 = 7

∴ The other end of the diameter B = (1,7)

(c) Given x^{2} + y^{2} – 6x + 2y – 31 = 0, A = (7,4), B = (x_{2}, y_{2}) = ?

Here C = (3,-1)

⇒ 7 + x_{2} = 6 & 4 + y_{2} = -2

⇒ x_{2} = -1 y_{2} = -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 x^{2} + y^{2} + 4x – 4y = 2. Find the value of a and b.

Answer:

Given (x_{1} y_{1}) = (a, b) (x_{2}, y_{2}) = (-5, 1) & Center of the given circle = (-2,2).

Equation of the circle when ends of diameter is given is

(x – x_{1}) (x – x_{2}) + (y – y_{1}) (y – y_{2}) = 0.

(x – a) (x + 5) + (y – b) (y – 1) = 0.

⇒ x^{2} – 5a – ax + 5x + y^{2} – by – y + b = 0.

x^{2} + y^{2} + x (5 – a) – y(b + 1) + (b – 5a) = 0 comparing.

This by given circle x^{2} + y^{3} + 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 x^{2} + y^{2} – 4x – 8y + k = 0 represents a point circle find k

Answer:

Given x^{2} + y^{2} – 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 x^{2} + y^{2} + ax + 2y + 5 = 0 represent a point circle find a

Answer:

Given x^{2} + y^{2} + 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 x^{2} + y^{2} + ax + by = 3 represents a circle with centre at (1, -3), find a and b.

Answer:

Given x^{2} + y^{2} + 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 x^{2} + y^{2} – 8x + 4y = 8.

Answer:

Here unit circle ⇒ r= 1

& Center of concentric circle x^{2} + y^{2} – 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 x^{2} + y^{2} – 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 x^{2} + y^{2} + 6x + 2y + 1 = 0, and passing through the point (-2, -3).

Answer:

Given x^{2} + y^{2} + 6x + 2y + 1 = 0, P = (-2,-3)

Centre = C(-3,-1),

Let the equations of the required circle is x^{2} + y^{2} + 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

x^{2} + y^{2} + 6x + 2y + 9 = 0.

Question 2.

Find the equation of the circle whose centre is same as the centre of the circle x^{2} + y^{2} – 6x + 4y + 9 = 0 and Passsing through the point (-2,3)

Answer:

Given x^{2} + y^{2} – 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

x^{2} + y^{2} – 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 x^{2} + y^{2} – 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 x^{2} + y^{2} – 6x + 4y+9=0.

Answer:

Given c= (-2,3) & given x^{2} + y^{2} – 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} = \((\sqrt{50})^{2}\)

x^{2} + y^{2} + 4x – 6y – 37 = 0.

Question 4.

Find the equation of the circle passing through the centre of the circle x^{2} + y^{2} – 2x – 4y – 20 = 0 and centre at (4, -2).

Answer:

Given x^{2} + y^{2} – 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 x^{2} + y^{2} – 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 x^{2} + y^{2} – 4x + 2y – 1 = 0.

Answer:

Given x^{2} + y^{2} – 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} = \((\sqrt{34})^{2}\)

x^{2} + 1 + 2x + y^{2} + 16 – 8y = 34

x^{2} + y^{2} + 2x – 8y – 17 = 0.

Question 6.

Find the equation of the circle concentric with the centre of circle x^{2} + y^{2} – 2x + 2y – 1 = 0 and having double its area.

Answer:

Given x^{2} + y^{2} – 2x + 2y – 1 = 0

c = (1, -1) & r_{ı} = \(\sqrt{1+1-(-1)}=\sqrt{3}\)

Given A_{2}, = 2A,

r_{2}^{2} =2 . \((\sqrt{3})^{2}\) = 2.3 = 6 ⇒ r_{2} = \(\sqrt{6}\)

Equation of the required circle with c = (1,-1) & r = is (x – 1)^{2} + (y + 1)^{2} = \((\sqrt{6})^{2}\).

x^{2} + 1 – 2x + y^{2} + 1 + 2y = 6 ⇒ x^{2} + y^{2} – 2x + 2y – 4 = 0.

Question 7.

Find the equation of the circle concentric with the centre of the circle 3x^{2} + 3y^{2} – 6x + 9y – 2 = 0 and \(\frac { 2 }{ 3 }\) having of its area.

Answer:

Given 3x^{2} + 3y^{2} – 6x + 9y – 2 = 0, divide by 3

x^{2} + y^{2} – 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 A_{2} = \(\frac{2}{3}\) A_{1}

πr_{2}^{2} = \(\frac{2}{3}\) . πr_{2}^{2} ⇒ r_{2}^{2} = \(\frac{2}{3} \times \frac{47}{12}=\frac{47}{18}\)

∴ Required circle with centre \(\left(1,-\frac{3}{2}\right), r_{2}-\sqrt{\frac{47}{18}} \text { is }(x-1)^{2}+\left(y+\frac{3}{2}\right)^{2}=(\sqrt{\frac{47}{18}})^{2}\)

x^{2} + 1 – 2x + y^{2} + 3y + \(\frac{9}{4}=\frac{47}{18}\)

x^{2} + y^{2} – 2x + 3y + \(\frac{23}{36}\) = 0 __________ × 36

36x^{2} + 36y^{2} – 72x + 108y + 23 = 0.

Question 8.

Find the equation of the diameter of the circle x^{2} + y^{2} + 6x – 2y = 6 which when produced passes through the point (1, -2).

Answer:

Given x^{2} + y^{2} + 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 – y_{1} = m(x – x_{1})

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 2x^{2} + 2y^{2} + 3x – 5y – 1 = 0 which when produced passes through the point (-1,2).

Answer:

Given 2x^{2} + 2y^{2} + 3x – 5y – 1 = 0 & p = (-1,2)

⇒ x^{2} + y^{2} + \(\frac { 3 }{ 2 }\)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 – y_{1}, = m(x – x_{1}); 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 x^{2} + y^{2} – 8x + 2y=0.

Answer:

Given x^{2} + y^{2} – 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 x^{2} + y^{2} – 6x + 15y – 16 = 0 intercepted by the x-axis.

Answer:

Given x^{2} + y^{2} – 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 x^{2 }+ y^{2} + 3x – y – 6 = 0 intercepted by the y-axis.

Answer:

Given x^{2} + y^{2} + 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 x^{2} + y^{2} – 6x – 4y – 12 = 0 on the coordinate axes.

Answer:

Given x^{2} + y^{2} – 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 x^{2} + y^{2} + 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 x^{2} + y^{2} ± 4x = 0.

(b) Let the general equation of the circle be x^{2} + y^{2} + 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 = g^{2} – C.

⇒ g^{2} – 25 = c. ….(3)

Solving 1, 2 & 3 we get g^{2} – 25 + 4g + 13 = 0

g^{2} + 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 x^{2} + y^{2} – 12x + 11 = 0

x^{2} + y^{2} + 4x – 21 = 0.

(C) Let the required equation of the circle is x^{2} + y^{2} + 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) 3^{2} + 4^{2} + 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

⇒ 2x^{2} + 2y^{2} – x – 47 = 0.

(d) Let required equation of the circle is x^{2} + y^{2} + 2gx + 2fy + c = 0

It passes through the points (1,2) & (2, 1) & centre lies on y-axis ⇒ g = 0 ….(1)

(1,2) 1^{2} + 2^{2} + 2g(1) + 2f(2) + c = 0 ⇒ 4f + C + 5 = 0 ….(2)

(2, 1) 2^{2} + 1^{2} + 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 x^{2} + y^{2} – 5 = 0

(e) Let required equation of the circle is x^{2} + y^{2} + 2gx + 2fy + c = 0

It passes through (0,5) & (6, 1)

(0,5) 0 + 5^{2} + 2g(0) + 2f(5) + c = 0

1f + C + 25 = 0 …(1)

(5, 1) 6^{2} + 1^{2} + 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 eq^{n} (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

x^{2} + y^{2} + 2\(\left(-\frac{5}{3}\right)\)x + 2(-1)y – 15 = 0 ⇒ × 3

3x^{2} + 3y^{2} – 10x – 6y – 45 = 0

(f) Let the required equation of the circle is x^{2} + y^{2} + 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) 1^{2} + (-4)^{2} + 2g(1) + 2f(-4) + c = 0

2g – 8f + C + 17 = 0 ….. (2)

⇒ (5,2) 5^{2} + 2^{2} + 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

x^{2} + y^{2} + 2 (3)x + 2 (-3)y – 47 = 0

x^{2} + y^{2} + 6x – 6y – 47 = 0

(g) Let the equation of required circle is x^{2} + y^{2} + 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

x^{2} + y^{2} + 2 (3) x + 2 (-1)y – 15 = 0

x^{2} + y^{2} + 6x – 2y – 15 = 0

(h) Let the equation of the required circle is x^{2} + y^{2} + 2gx + 2fy + c = 0

If passes through (1, 1) & (2,2)

(1,1) 1^{2} + 1^{2} + 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 ⇒ g^{2} + f^{2} – 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

g^{2} + f^{2} + 2g + 2f = -1

(-3 -f)^{2} + f^{2} + 2 (-3-f) + 2f + 1 = 0

2f^{2} + 6f + 4 = 0 ÷ 2

f^{2} + 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 = g^{2}+ f^{2} -1 = 1 + 4 – 1 = 4

∴ Required circles are x^{2} + y^{2} – 2x – 4y + 4 = 0 and x^{2} + y^{2} – 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 x^{2} + y^{2} + 2gx + 2fy + c = 0

This equation passes through (*0,2) (3,0) & (3,2)

(0,2) 0^{2} + 2^{2} + 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 x^{2} + y^{2} + 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 x^{2} + y^{2} + 4x + 6y – 12 = 0

(c) (1, 1) (5,-5) (6,-4)

The general equation of the circle

x^{2} + y^{2} + 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

x^{2} + y^{2} + \(\frac { 222x }{ 21 }\) + \(\frac { 20y }{ 21 }\) – \(\frac { 284 }{ 21 }\) = 0 – × 21

21x^{2} + 21y^{2} + 222x + 20y – 284 = 0

(d) Let the general eqn, x^{2} + y^{2} + 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

∴ 2x^{2} + 2y^{2} – 8x – 7y + 6 = 0.

(e) Let the general eqn. x^{2} + y^{2} + 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 x^{2} + y^{2} – 4x – 6y – 12 = 0.

(f) (p, q) (p, 0) (0,9)

Let the general eqn. x^{2} + y^{2} + 2gx + 2fy + c = 0.

The circle passes through the points

(p, q) (p, 0) (0,q)

(p, q) p^{2} + q^{2} + 2pq + 2qf + c = 0 … (1)

(p, 0) p^{2} + 2pg + c = 0 …… (2)

(0,q) q^{2} + 2qf + c= 0 … (3)

1 – 2 gives 2qf = -q^{2} ⇒ f = \(-\frac{q}{2}\)

1 – 3 gives 2pg = -p^{2} ⇒ g= \(-\frac{p}{2}\) & c = 0

∴ Required eqn. is x^{2} + y^{2} – px – qy = 0.

(g) Let the general eqn, x^{2} + y^{2} + 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 x^{2} + y^{2} + \(\frac{2}{3}\)x – \(\frac{20}{3}\)y + \(\frac{17}{3}\) = 0

3x^{2} + y^{2} + 2x – 20y + 17 = 0

(h) Let the general equation x^{2} + y^{2} + 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 = -a^{2} ⇒ g = \(-\frac{a}{2}\)

(0, b) 2bf + c = -b^{2} ⇒ f = \(-\frac{b}{2}\)

∴ Required equation is x^{2} + y^{2} – 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 x^{2} + y^{2} + 2gx + 2fy + c = 0 is required equation

(3,0) 6g + C + 9 = 0 … (1)

(-3,0) -6g + C + 9 = 0 …. (2)

r = 5; 25 = g^{2} + f^{2} -C

Adding 1 and 2 we get c = -9 and g = 0 and f = ±4

∴ The required equation is x^{2} + y^{2} + 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 ⇒ r^{2} = g^{2} + f^{2} -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

x^{2} + y^{2} -14x – 12y + 76 = 0

x^{2} + y^{2} – 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) a^{2} + 2ga = 0 ⇒ g = \(-\frac{a}{2}\)

(0, -b) b^{2} + 2fb = 0 ⇒ f = \(-\frac{b}{2}\)

∴ The required equation is x^{2} + y^{2} – 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 x^{2} + y^{2} + 2gx + 2fy + c = 0 passes through origin ⇒ c = 0 …. (1)

(a,0) ⇒ a^{2} + 2ga = 0 = g ⇒ \(-\frac{a}{2}\) .. (2)

(0, -b) ⇒ b^{2} – 2fb = 0 = f ⇒ \(-\frac{b}{2}\) … (3)

∴ The required equation is x^{2} + y^{2} – ax + by = 0.

Question 7.

Find the length of the chord intercepted by the

(a) circle x^{2} + y^{2} – 8x – 6y = 0 and the line x – 7y – 8 = 0

(b) circle x^{2} + y^{2} = 9 and the line x + 2y = 3

(c) circle x^{2} + y^{2} – 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) x^{2} + y^{2} = 9 and the line x + 2y = 3.

(b) x^{2} + y^{2} – 6x – 2y + 5 = 0 and the line x – y + 1=0.

(c) x^{2} + y^{2} + 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 x^{2} + y^{2} = 9

(3 – 2y)^{2} + y^{2} = 9

5y^{2} – 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 x^{2} + y^{2} – 6x – 2y + 5 = 0

x^{2} + (x + 1)^{2} – 6x – 2 (x + 1) + 5 = 0

2x^{2} – 6x + 4 = 0

x^{2} – 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 x^{2} + y^{2} + 4x + 6y – 12 = 0

(6 – 4y)^{2} + y^{2} + 4 (6 – 4y) + 6y – 12 = 0

17y^{2} – 58y + 48 = 0

17y^{2} – 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 x^{2} + y^{2} – 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 x^{2} + y^{2} – 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 x^{2} + y^{2} + \(\frac{246}{50} x\) + \(\frac{147}{25} y\) + \(\frac{196}{25} \) = 0

25x^{2} + 25y^{2} + 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