Students can Download Basic Maths Question Bank Chapter 15 Circles Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

## Karnataka 2nd PUC Basic Maths Question Bank Chapter 15 Circles

### 2nd PUC Basic Maths Circles One Mark Questions and Answers

Question 1.

Find the equation of the circle with centre at orgin and radius 4 units.

Answer:

x^{2} + y^{2} = 4^{2}

x^{2} + y^{2} = 16

Question 2.

Find the equation of the circle with centre at (-3,2) and radius 5 units.

Answer:

(x + 3)^{2} + (y – 2)^{2} = 5^{2}.

Question 3.

Write the equation of the circle with centreat (1,1) and radius 2 units

Answer:

(x – 1)^{2} + (y – 1)^{2} = (√2)^{2}

(x – 1)^{2} + (y – 1)^{2} = 2

Question 4.

Find the equation of the point circle with centre at (a) (4, -5), (b) (-3,2), (c) (1,0)

Answer:

(a) (x – 4)^{2} + (y + 5)^{2} = 0

(b) (x + 3)^{2} + (y – 2)^{2} = 0

(c) (x – 1)^{2} + y^{2} = 0

Question 5.

Find the centre of the circle x^{2} + y^{2} – 4x – 6y + 1 = 0.

Answer:

Centre = (2, 3)

Question 6.

Find the radius of the circle x^{2} + y^{2} + 4x – 2y – 4 = 0.

Answer:

g = 2, f = – 1, C = – 4.

r =\(\sqrt{g^{2}+f^{2}-C}\) =\(\sqrt{4+1+4}\) = √9 = 3.

Question 7.

Find the centre of the circle 4X^{2} + 4y^{2} – 15x – 18y + 11 = 0.

Answer:

Divide the circle of by 4

x^{2} + y^{2} – \(\frac{15}{4}\)x – \(\frac { 18 }{ 4 }\)y + \(\frac { 11 }{ 4 }\) = 0

Centre = \(\left(\frac{15}{8}, \frac{9}{4}\right)\)

Question 8.

Find the equation of the circle having the centre at (3,4) and touching the x-axis.

Answer:

Given g = – 3, f=- 4 and c = g^{2} = 9 as ittouchs x-axis.

∴ Eqn. of the circle is x^{2} + y^{2} + 2gx + 2fy + c = 0

x^{2} + y^{2} – 6x – 8y + 9 = 0.

Question 9.

Find the length of the chord of the circle x^{2} + y^{2} – 6x + 15y -16 = 0 intercepted by the x- axis.

Answer:

Length of the chord intercepted by x-axis is 2\(\sqrt{g^{2}-c}\)

= 2\(\sqrt{9-(-16)}\) = 2\(\sqrt{25}\) = 2.5 = 10.

Question 10.

Find the equation of a circle whose centre is (4,2) and touchs the y-axis.

Answer:

If the circle touchess the y-axis then f^{2} = c

The eqn. of the circle is x^{2} + y^{2} + 2gx + 2fy + c = 0

With g = – 4, f = – 2, c = f^{2} = 4 is x^{2} + y^{2} – 8x – 4y + 4 = 0.

Question 11.

Find the length of the chord of the circle x^{2} + y^{2} + 3x – 2 = 0 intercepted by y-axis.

Answer:

Given f = 0, c = – 2

The length of the chord intercepted by y-axis is given by = 2\(\sqrt{f^{2}-c}=2 \sqrt{0-(-2)}=2 \sqrt{2}\)

Question 12.

Write the unit circle with centre at the origin in.

Answer:

x^{2} + y^{2} = 1.

### 2nd PUC Basic Maths Circles Two Marks Questions and Answers

Question 1.

Find the centre and radius of the circle x^{2} + y^{2} + 4x + 2y – 1 = 0.

Answer:

Centre = (- g, -f) = (-2, -1), c = – 1

r = \(\sqrt{g^{2}+f^{2}-c}=\sqrt{4+1-(-1)}=\sqrt{5+1}=\sqrt{6}\)

Question 2

Find the equation of the circle with A(x_{1}, y_{1}) and B(x_{2}, y_{2}) as the ends of the diameter.

Answer:

A(x_{1}, y_{1}) and B(x_{2}, y_{2}) be 2 given points

P(x, y) be any point on the circle join

AP and BP and AP is perpendicular to BP

∴∠APB = 90° (Angle in a semicircle)

∴ Slope of AB x slope of PB = – 1

\(\frac{y-y_{1}}{x-x_{1}} \times \frac{y-y_{2}}{x-x_{2}}\) = -1 ⇒ (x – x_{1}) (x – x_{2}) + (y – y_{1})(y – y_{2}) = 0 is the required equation.

Question 3.

Find the equation of the circle whose ends of diameter are (3,1) and (-4,2).

Answer:

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

given (X_{1}, y_{1}) = (3, 1), (x_{2}, y_{2}) = (-4,2)

∴ (x – 3)(x + 4) + (y – 1)(y – 2) = 0

⇒ x^{2} + y^{2} + x – 3y -10 = 0 is the required equation.

Question 4.

Find the equation of the circle with centre (4,3) and which passes through (0,0).

Answer:

Given centre = (4, 3)

and r = distance between (4, 3) and (0, 0)

r = \(\sqrt{(4-0)^{2}+(3-0)^{2}}\) = \(\sqrt{16+9}\) = 5

∴The required equation of the circle with centre (4, 3) and radius = 5 units is

(x – 4)^{2} + (y – 3)^{2} = 5^{2}

x^{2} + y^{2} – 8x – 6y = 0.

Question 5.

Find the equation of the circle whose two diameters are x + 2y = 3 and x – y = 6 and radius is 6 units.

Answer:

Solving the 2 diameters we get the centre.

∴The centre = (5, -1)andr = 6

Equation of circle is

(x – 5)^{2} + (y + 1)^{2} = 6^{2}

or x^{2} + y^{2} – 10x + 2y – 10 = 0.

Question 6.

If one end of the diameter of the circle x^{2} + y^{2} – 2x – 4y + 4 = 0 is (1,1). Find the other end.

Answer:

Centre of the given circle = (1,2)

One end is given as A (1, 1)

Let the other end be B(x, y) = ?

W.K.T. centre = mid point of the diameter

∴(1,2) = \(\left(\frac{1+x}{2}, \frac{1+y}{2}\right)\)

⇒ 1 + x = 2 1 + y = 4

⇒ x = 1 y = 3

∴The other end of the diameter = B(l, 3).

Question 7.

Find the equation to the circle whose centre is same as the centre of the circle 2x^{2} + 2y^{2} – 6x + 8y – 3 = 0 and whose radius is 3 units.

Answer:

Given circle 2x^{2} + 2y^{2} – 6x + 8y – 3 = 0

x^{2} + y^{2} – 3x + 4y – \(\frac { 3 }{ 2 }\) = 0 .

The equation of the circle whose centre is same as the centre of x^{2} + y^{2} – 3x + 4y – \(\frac { 3 }{ 2 }\) = 0 is of the form x^{2} + y^{2} – 3x + 4y + c = 0

g = \(\frac { -3 }{ 2 }\), f = 2, c = \(\frac { -3 }{ 2 }\)

r = \(\sqrt{g^{2}+f^{2}-c}\)

S.B.S 3 = \(\sqrt{\frac{9}{4}+4-c}\)

9 = \(\frac{9+16-4 c}{4}\)

– 4c + 25 = 36

– 4c = 11 ⇒ C = \(\frac { 11 }{ -4 }\)

∴Required circle is x^{2} + y^{2} – 3x + 4y – \(\frac { 11 }{ 4 }\)

⇒ x^{2} + y^{2} – 3x + 4y – 11 =0.

Question 8.

Find the equation of the circle passing through the points (1,0) (0,1) and (0,0).

Answer:

Let the required circle is x^{2} + y^{2} + 2gx + 2fy + c = 0

This equation passing through (0, 0) (1, 0), (0, 1)

∴(0, 0) ⇒ c = 0 ‘

(1,0) ⇒ 1 – 2g = 0 ⇒ g = \(-\frac{1}{2}\)

(0, 1) ⇒ 1 + 2f = 0 ⇒ f = \(-\frac{1}{2}\)

∴ The equation of the circle is

x^{2} + y^{2} – x – y = 0.

Question 9.

Find the equation of the circle whose centre is same as the circle x^{2} + y^{2} – 2x + 4y -11 = 0 and the radius is 4 units.

Answer:

Centre of the given circle is (1, -2)

∴The required circle with centre (1, -2) and r = 4 units is

(x – 1)^{2} + (y + 2)^{2} = 4^{2}

⇒ x^{2} + y^{2} – 2x + 4y – 11 = 0.

Question 10.

If (a, b) and (-5,1) are the ends of a diameter of the circle x^{2} + y^{2} + 4x – 4y – 2 = 0 find a ‘ and b.

Answer:

Centre of the given circle = (- 2, 2)

W.k.t. centre = mid point of the diameter

(-2,2) = \(\left(\frac{a-5}{2}, \frac{b+1}{2}\right)\)

a – 5 = -4, b + 1=4

a = 1, b = 3

∴ (a, b) = (1, 3).

Question 11.

S.T. the line (3x – 4y + 6) = 0 touches the circle x^{2} + y^{2} – 6x + 10y – 15 = 0.

Answer:

Centre of the given circle ~ (3,-5)

and r = \(\sqrt{g^{2}+f^{2}-c}=\sqrt{9+25-(-15)}=\sqrt{49}=7\)

Length of the perpendicular from the centre (3, -5) to the line 3x – 4y + 6 = 0 is

Lt. of perpendicular = radius = 7

∴The line touches the circle.

Question 12.

If the radius of the circle x^{2} + y^{2} – 2x + 3y + k = 0 is \(\frac { 5 }{ 2 }\) find k.

Answer:

Centre of the given circle = \(\left(+1, \frac{-3}{2}\right)\)

Question 13.

Find the value of k such that the line 3x – 4y + k = 0 may be a tangent to the circle x^{2} + y^{2} = 25

Answer:

Centre of the circle = (0, 0)

radius = 5

Since the line is a tangent to the circle we get

radius = Length of the perpendicular from the centre (0, 0)

Question 14.

Find A, if the line 3x + y + λ = 0 touches the circle x^{2} + y^{2} – 2x – 4y – 5 = 0.

Answer:

Centre = (1, 2) and r = \(\sqrt{1+4+5}=\sqrt{10}\)

Since the line touches the circle ∴

radius = Length of the perpendicular.

Question 15.

Find the centre of the circle, two of the diameters are x + y = 2 and x – y = 0.

Answer:

Point of intersection of the diameters is the centre

∴Solving the 2 diameters we get

∴Center = (1,1)

Question 16.

Find the equation of the circle two of the diameters x + y = 4 and x – y = 2 and passing through the point (2, -1).

Answer:

Point of intersection of two diameters is centre

y = 4 – x

= 4 – 3

= 1

∴The centre = (3,1).

radius = distance between the centre (3,1) and the point (2, -1)

r = \(\sqrt{(3-2)^{2}+(1+1)^{2}}\)

\(=\sqrt{1+4}=\sqrt{5}\)

∴Equation of the required circle with centre (3, 1) and r = √5 units is

(x – 3)^{2} + (y – 1)^{2} = (√5)

⇒ x^{2} + y^{2} – 6x – 2y + 5 = 0.

Question 17.

S.T. the circle x^{2} + y^{2} + 4x – 3y + 4 = 0 touches x-axis.

Answer:

g = 2, c = 4

Condition for the circle to touch x-axis is g^{2} = c

Here 2^{2} = 4 ⇒ 4 = 4

∴ The given circle touches x-axis.

Question 18.

S.T. the circle x^{2} + y^{2} – 3x + 8y + 16 = 0 touches y-axis.

Answer:

Here f = 4, c = 16

The condition for the circle to touch y-axis is f^{2} = c

Here 4^{2} = 16 ⇒ 16 = 16

∴ The given circle touches y-axis.

Question 19.

S.T. the circle x^{2} + y^{2} – 2x + 2y + 1 = 0 touches both the co-ordinate axes.

Answer:

The condition for the circle to touch both the axes.

is g^{2} = f^{2} = c

Here g = 1 f = – 1 and c = 1 .

∴ (1)^{2} = (-1)^{2} = 1 ⇒ 1 = 1 = 1

Hence the given circle touches both the axes.

Question 20.

Find the equation of the tangents to the circle x^{2} + y^{2} + 2x + 4y – 4 = 0 which are parallel to the line 5x + 12y + 6 = 0

Answer:

Centre of the given circle = (-1, -2) and r = \(\sqrt{1+4+4}=\sqrt{9}\) = 3

Equation of the tangent parallel to the given line can be taken as 5x + 12y + k = 0.

Length of parpendicur from (-1, -2) to the line 5x + 12y + k = 0

∴ k – 29 = ±39 ⇒ k = 68 or -10

∴The equations of tangents are 5x + 12y + 68 = 0 and

5x + 12y – 10 = 0.

Question 21.

Find the equation of the tangents to the circle x^{2} + y^{2} – 2x – 4y + 1 = 0 which are perpendicur to the line 3x – 4y – 7 = 0.

Answer:

Centre (1,2), r = \(\sqrt{1+4-1}=\sqrt{4}\) = 2

Any line ⊥^{lar} to the given line can be .taken as 4x + 3y + k = 0

∴Length of the ⊥^{lar} from (1,2) to the line 4x + 3y + k = 0 is

⇒ 10 + k ± 10

⇒ k = 0 or – 20

∴The required equations are 4x + 3y = 0 and 4x + 3y – 20 = 0.

Question 22.

S.T. the line 1x + y = 4 passes through the centre of the circle x^{2} + y^{2} – 3x – 2y + 2 = 0.

Answer:

Centre = \(\left(\frac{3}{2}, 1\right)\)

Put x = \(\frac{3}{2}\), y = 1 in the equation 2x + y = 4

4 = 4

Thus the center \(\left(\frac{3}{2}, 1\right)\) lies on the line 2x + y = 4.

### 2nd PUC Basic Maths Circles Five or Six Marks Questions and Answers

Question 1.

Find the equation of the circle passing through the points

(i) (2,0) (-1, 3) and (-2, 0).

Answer:

Let the equation of the required circle is

x^{2} + y^{2} + 2gx + 2fy + c = 0 ‘

Given that this equation passes through the points.

(2, 0) ⇒ 4 + 0 + 4g + 0 + c = 0

4g + 4 + c = 0 _____ (1)

(-1,3) ⇒ 1 + 9 – 2g + 6f + c = 0

-2g + 6f + c + 10 = 0 ______ (2)

(-2,0) ⇒ 4 + 0 – 4g + c = 0

-4g + 4 + c = 0 (3)

Solving equations 1, 2 and 3 we get

put Eqn(3) in Eqn(l) i.e., 4 + c = 4g

we get c = – 4, g = 0, and f= – 1

∴ The required equation of the circle is x^{2} + y^{2} – 2y – 4 = 0.

Question 2.

(1,1) (5,-5) and (6,-4).

Answer:

Let the equation of the required circle is

x^{2} + y^{2} + 2gx + 2fy + c = 0

Given that this equation passes through the points.

(1,1) ⇒ 1 + 1 + 2g + 2f + c

2g + 2f + c + 2 = 0 — (1)

(5,-5) ⇒ 25 + 25 + 10g – 10f + c = 0

10g – 10f + c + 50 = 0 (2)

(6, -4) ⇒ 36 + 16 + 12g – 8f + c = 0

I2g – 8f + c + 52 = 0 (3)

Solving equations 1, 2 and 3 we get

g = -3, c = 0, f = 2

∴The required equation of the circle is x^{2} + y^{2} – 6y + 4y = 0.

Question 3.

Find the equation of a circle passing through (0,0) (0,2) and (- 3, -1).

Answer:

Let the required equation of the circle be

x^{2} + y^{2} + 2gx + 2fy + c = 0

This equation passes through the points.

(0, 0) ⇒ 0 + 0 + 0 + c – 0 ⇒ c = 0 ______ (1)

(0, 2) ⇒ 0 + 4 + 0 + 4f + c = 0 ⇒ 4f + 4 = 0 ______ (2)

f = -1

(-3, -1) ⇒ 9 + 1 – 6g – 2f + c = 0 ⇒ -6g + 2 + 10 = 0

– 6g = – 12

g = 2 ______ (3)

∴Required equation of the circle is x^{2} + y^{2} + 4x – 2y = 0.

Question 4.

A circle has its centre on the y-axis and passess through (- 1, 3) and (2, 5). Find its equation.

Answer:

Let the required equation of the circle be

x^{2} + y^{2} + 2gx + 2fy + c = 0

Given the the centre (- g, – f) lies on the y-axis.

∴x co-ordinate of the centre is 0 ⇒ g = 0 _______ (1)

Also given the required equation passess through the points (-1,3) and (2, 5)

(-1,3) ⇒ (-1)^{2} + (3)^{2} + 2(-1)g + 2(3)f + c = 0

-2g + 6f + c + 10 = 0 (g = 0)

∴6f + c + 10 = 0 _______ (2)

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

4g + 10f + 29 + c = 0

∴10f + 29 + c = 0 _______ (3)

Solving equations 2 and 3 we get f = \(-\frac{19}{4}\) & c = \(\frac { 37 }{ 2 }\)

we get g = 0, f = \(\frac{19}{4}\), c = \(-\frac{37}{2}\)

∴Required equation of the circle is

2x^{2} + 2y^{2} – 19y + 37 = 0.

Question 5.

Find the equation of a circle passing through (0,0) (1,1) and has its centre on x-axis.

Answer:

Let the required equation of the circle is

x^{2} + y^{2} + 2gx + 2fy + c = 0

Given the the centre (- g, -f) lies on the x-axis ⇒ f = 0 _______ (1)

Also this equation passess through the points (0, 0) and (1,1)

(0,0) ⇒ c = 0 _______ (2)

(1, 1) ⇒ 1 + 1 + 2g = 0 ⇒ g = -1 _______ (3)

The required equation of the circle is x^{2} + y^{2} – 2x = 0.

Question 6.

Find the equation of the circle passing through the points (0,4) and (4,7) and its centre lies on the line 3x + 4y = 7.

Answer:

Le^{2} + y^{2} + 2gx + 2fy + c = 0

Given this circle passess through the points (0, 4) and (4, 7)

(0,4) ⇒ 16 + 8f + c = 0 _______ (1)

(4,7) ⇒ 8g + 14f + c = – 65 _______ (2)

Also the centre (- g, – f) lies on the line 3x + 4y = 7

∴ – 3g – 4f = 7 _______ (3)

Solving equations 1, 2, and 3 we get

g = – 11, f = \(\frac { 13 }{ 2 }\), c = -68

The required equations of the c ircle is

∴x^{2} + y^{2} – 22x + 13y – 68 = 0.

Question 7.

S.T. the line 3x – 4y – 20 = 0 touches the circle x^{2} + y^{2} – 2x- 4y — 20 = 0 and also find the point contact.

Answer:

The centre of the given circle = (1,2) and r = \(\sqrt{1^{2}+2^{2}-(-20)}=\sqrt{1+4+20}=\sqrt{25}\) = 5

Condition is

Length of the ⊥^{lar} from the centre (1, 2) to the line = radius

5 = 5

∴ The line 3x – 4y – 20 = 0 touches the circle

Any line perpendicular to 3x – 4y – 20 = 0 is of the from

4x + 3y + k = 0 but it passess through (1,2)

4 + 6 + k = 0 ⇒ k = — 10

∴ Equation of the ⊥^{lar} line is 4x + 3y- 10 = 0

The point of interesection of the lines 4x + 3y – 10 = 0 and 3x – 4y -20 = 0.

Solving the above equations we get

x = 4 and y = – 2

∴ The point of contact is (4, -2).

Question 8.

Find the equation of the circle which passess through (2, 3). having its centre on the je-axis and radius 5 units.

Answer:

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

This passess through the point (2, 3) we get

4 + 9 + 4g + 6f + c = 0

4g + 6f + c + 13 = 0 _____ (1)

Centre (- g, -f) lies on the x axis ⇒ f = 0 _____ (2)

Also given r = 5 units

\(\sqrt{g^{2}+f^{2}-c}=5\) = 5 ⇒ g^{2} + f^{2} – c = 25

g^{2} – c = 25 _____ (3)

from eqn.(l) we get .

4g + 13 = – c

Eqn(3) gives.

g^{2} + 4g + 13 = 25

g^{2} + 4g – 12 = 0

(g + 6)(g – 2) = 0

g = – 6 or 2

when g = – 6, c = 11

when g = 2 c = – 21

∴ The required equations of the circles are

x^{2} + y^{2} – 12x + 11 = 0

x^{2} + y^{2} + 4x – 21 = 0.

Question 9.

S.T. the points (1,1) (- 2,2) (- 6,0) and (- 2, – 8) are concyclic.

Answer:

First find the equation of the circle passess through the points (1, 1) (-2, 2) and (- 6, 0). If the 4th point (-2, -8) satisfies the circle then the points are concyclic

Let equation of the required circle is

x^{2} + y^{2} + 2gx + 2fy + c = 0

This equation passess through the point (1, 1) (-2, 2) and (-6, 0)

(1, 1) ⇒ 2g + 2f + c + 2 = 0 _____ (1)

(-2, 2) ⇒ 4g + 4f + c = -8 _____ (2)

(-6,0) ⇒ -12g + c = – 36 _____ (3)

Eqn. 1 – 2 gives 3g – f = 3 _____ (4)

Eqn. 2 – 3 gives 2g + f = 7 _____ (5)

Adding eqns. 4 and 5 we get

5g = 10 ⇒ g = 2

f = 7 – 2g = 7 – 4 = 3

f = 3

2g + 2f + c = -2

c = – 2 – 2g – 2f

= – 2 – 4 – 6

c = – 12

The equation of the circle is x^{2} + y^{2} + 4x + 6y – 12 = 0

put x = – 2, and y = – 8 in the above circle ‘

we get (-2)^{2} + (-8)^{2} + 4(-2) + 6(—8) -12 = 0

4 + 64 – 8 – 48 – 12 = 0

68 – 68 = 0

0 = 0

∴ The 4 points are concyclic.

Question 10.

S.T. the following points arc concyclic (0,0) (1,1) (5, -5) and (6, -4).

Answer:

Let the equation of the circle passing through

(0, 0) (1, 1) and (5,-5) be x^{2} + y^{2} + 2gx + 2fy + c = 0

It passes through (0,0) ⇒ c = 0 _____ (1)

It passes through (1, 1) ⇒ 2g + 2f + c + 2 = 0

2g + 2f + 2 = 0

g + f + 1 = 0 _____ (2)

It passes through (5,-5) ⇒ 10g – 10f + 50 = 0

g – f + 5 = 0 _____ (3)

Solving 2 and 3 we get g = – 3 and f = 2

Thus, the equation of the circle = x^{2} + y^{2} – 6x + 4y = 0

Put x = 6 and y = – 4 in L.H.S of the above eqn.

36 + 16 – 36 – 16 = 0

0 = 0

Thus, the point (6, -4) lies on the circle.

Hence, the given 4 points lie on the circle and concyclic.

Question 11.

Find the length of the chord intercepted by the circle x^{2} + y^{2} – 8x – 6y = 0 and the line x – 7y – 8 = 0

Answer:

Centre of the given circle = (4, 3)

and r = \(\sqrt{16+9-0}\) = 5

P = Length of ⊥^{lar} from C (4, 3) to the line x – 7y – 8 = 0

Length of the chord

∴ The length of the chord = 5√2

Question 12.

Find the length of the chord intersepted by the the circle x^{2} + y^{2} = 9 and the line x + 2y = 3.

Answer:

Centre of the given circle = (0, 0) and radius = 3

P = Length of the ⊥^{lar} from the centre (0,0) to the line x + 2y = 3