# 2nd PUC Basic Maths Question Bank Chapter 15 Circles

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## 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.
x2 + y2 = 42
x2 + y2 = 16

Question 2.
Find the equation of the circle with centre at (-3,2) and radius 5 units.
(x + 3)2 + (y – 2)2 = 52.

Question 3.
Write the equation of the circle with centreat (1,1) and radius 2 units
(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)
(a) (x – 4)2 + (y + 5)2 = 0
(b) (x + 3)2 + (y – 2)2 = 0
(c) (x – 1)2 + y2 = 0

Question 5.
Find the centre of the circle x2 + y2 – 4x – 6y + 1 = 0.
Centre = (2, 3)

Question 6.
Find the radius of the circle x2 + y2 + 4x – 2y – 4 = 0.
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 4X2 + 4y2 – 15x – 18y + 11 = 0.
Divide the circle of by 4
x2 + y2 – $$\frac{15}{4}$$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.
Given g = – 3, f=- 4 and c = g2 = 9 as ittouchs x-axis.
∴ Eqn. of the circle is x2 + y2 + 2gx + 2fy + c = 0
x2 + y2 – 6x – 8y + 9 = 0.

Question 9.
Find the length of the chord of the circle x2 + y2 – 6x + 15y -16 = 0 intercepted by the x- axis.
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.
If the circle touchess the y-axis then f2 = c
The eqn. of the circle is x2 + y2 + 2gx + 2fy + c = 0
With g = – 4, f = – 2, c = f2 = 4 is x2 + y2 – 8x – 4y + 4 = 0.

Question 11.
Find the length of the chord of the circle x2 + y2 + 3x – 2 = 0 intercepted by y-axis.
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.
x2 + y2 = 1.

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

Question 1.
Find the centre and radius of the circle x2 + y2 + 4x + 2y – 1 = 0.
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(x1, y1) and B(x2, y2) as the ends of the diameter.

A(x1, y1) and B(x2, y2) 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 – x1) (x – x2) + (y – y1)(y – y2) = 0 is the required equation.

Question 3.
Find the equation of the circle whose ends of diameter are (3,1) and (-4,2).
(x – x1) (x – x2) + (y – y1) (y – y2) = 0
given (X1, y1) = (3, 1), (x2, y2) = (-4,2)
∴ (x – 3)(x + 4) + (y – 1)(y – 2) = 0
⇒ x2 + y2 + 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).
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 = 52
x2 + y2 – 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.
Solving the 2 diameters we get the centre.

∴The centre = (5, -1)andr = 6
Equation of circle is
(x – 5)2 + (y + 1)2 = 62
or x2 + y2 – 10x + 2y – 10 = 0.

Question 6.
If one end of the diameter of the circle x2 + y2 – 2x – 4y + 4 = 0 is (1,1). Find the other end.
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 2x2 + 2y2 – 6x + 8y – 3 = 0 and whose radius is 3 units.
Given circle 2x2 + 2y2 – 6x + 8y – 3 = 0
x2 + y2 – 3x + 4y – $$\frac { 3 }{ 2 }$$ = 0 .
The equation of the circle whose centre is same as the centre of x2 + y2 – 3x + 4y – $$\frac { 3 }{ 2 }$$ = 0 is of the form x2 + y2 – 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 x2 + y2 – 3x + 4y – $$\frac { 11 }{ 4 }$$
⇒ x2 + y2 – 3x + 4y – 11 =0.

Question 8.
Find the equation of the circle passing through the points (1,0) (0,1) and (0,0).
Let the required circle is x2 + y2 + 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

x2 + y2 – x – y = 0.

Question 9.
Find the equation of the circle whose centre is same as the circle x2 + y2 – 2x + 4y -11 = 0 and the radius is 4 units.
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 = 42
⇒ x2 + y2 – 2x + 4y – 11 = 0.

Question 10.
If (a, b) and (-5,1) are the ends of a diameter of the circle x2 + y2 + 4x – 4y – 2 = 0 find a ‘ and b.
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 x2 + y2 – 6x + 10y – 15 = 0.
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 x2 + y2 – 2x + 3y + k = 0 is $$\frac { 5 }{ 2 }$$ find k.
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 x2 + y2 = 25
Centre of the circle = (0, 0)
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 x2 + y2 – 2x – 4y – 5 = 0.
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.
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).
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)
⇒ x2 + y2 – 6x – 2y + 5 = 0.

Question 17.
S.T. the circle x2 + y2 + 4x – 3y + 4 = 0 touches x-axis.
g = 2, c = 4
Condition for the circle to touch x-axis is g2 = c
Here 22 = 4 ⇒ 4 = 4
∴ The given circle touches x-axis.

Question 18.
S.T. the circle x2 + y2 – 3x + 8y + 16 = 0 touches y-axis.
Here f = 4, c = 16
The condition for the circle to touch y-axis is f2 = c
Here 42 = 16 ⇒ 16 = 16
∴ The given circle touches y-axis.

Question 19.
S.T. the circle x2 + y2 – 2x + 2y + 1 = 0 touches both the co-ordinate axes.
The condition for the circle to touch both the axes.
is g2 = f2 = 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 x2 + y2 + 2x + 4y – 4 = 0 which are parallel to the line 5x + 12y + 6 = 0
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 x2 + y2 – 2x – 4y + 1 = 0 which are perpendicur to the line 3x – 4y – 7 = 0.
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 x2 + y2 – 3x – 2y + 2 = 0.
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).
Let the equation of the required circle is
x2 + y2 + 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 x2 + y2 – 2y – 4 = 0.

Question 2.
(1,1) (5,-5) and (6,-4).
Let the equation of the required circle is
x2 + y2 + 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 x2 + y2 – 6y + 4y = 0.

Question 3.
Find the equation of a circle passing through (0,0) (0,2) and (- 3, -1).
Let the required equation of the circle be
x2 + y2 + 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 x2 + y2 + 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.
Let the required equation of the circle be
x2 + y2 + 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) ⇒ 22 + 52 + 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

2x2 + 2y2 – 19y + 37 = 0.

Question 5.
Find the equation of a circle passing through (0,0) (1,1) and has its centre on x-axis.
Let the required equation of the circle is
x2 + y2 + 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 x2 + y2 – 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.
Le2 + y2 + 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
∴x2 + y2 – 22x + 13y – 68 = 0.

Question 7.
S.T. the line 3x – 4y – 20 = 0 touches the circle x2 + y2 – 2x- 4y — 20 = 0 and also find the point contact.
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.
Let the required equation of the circle is x2 + y2 + 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 ⇒ g2 + f2 – c = 25
g2 – c = 25 _____ (3)
from eqn.(l) we get .
4g + 13 = – c
Eqn(3) gives.
g2 + 4g + 13 = 25
g2 + 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
x2 + y2 – 12x + 11 = 0
x2 + y2 + 4x – 21 = 0.

Question 9.
S.T. the points (1,1) (- 2,2) (- 6,0) and (- 2, – 8) are concyclic.
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
x2 + y2 + 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 x2 + y2 + 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).
Let the equation of the circle passing through
(0, 0) (1, 1) and (5,-5) be x2 + y2 + 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 = x2 + y2 – 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 x2 + y2 – 8x – 6y = 0 and the line x – 7y – 8 = 0
and r = $$\sqrt{16+9-0}$$ = 5