Circle

EXERCISE 24.2

QUESTION 1

Find the co-ordinates of the center and radius of each of the following circles:

(i) x^{2}+y^{2}+6 x-8 y-24=0

Sol :

The given equation can be rewritten as x^{2}+y^{2}+2(3) x-2(4) y-24=0

\therefore Center =(-3,4)

And, radius =\sqrt{(3)^{2}+(4)^{2}-(-24)}

=\sqrt{49}

= 7

 

(ii) 2 x^{2}+2 y^{2}-3 x+5 y=7

Sol :

The given equation can be rewritten as x^{2}+y^{2}-\frac{3 x}{2}+\frac{5 y}{2}-\frac{7}{2}=0

Center=\left(\dfrac{3}{4},\dfrac{-5}{\phantom{-}4}\right)

Radius=\sqrt{\left(\frac{3}{4}\right)^{2}+\left(\frac{-5}{4}\right)^{2}-\left(-\frac{7}{2}\right)}

=\sqrt{\frac{34+56}{16}}

=\sqrt{\frac{90}{16}}

=\frac{3 \sqrt{10}}{4}

 

(iii) 1 / 2\left(x^{2}+y^{2}\right)+x \cos \theta+y \sin \theta-4=0

Sol :

The given equation can be rewritten as x^{2}+y^{2}+2 x \cos \theta+2 y \sin \theta-8=0

Center =(-\cos \theta,-\sin \theta)

radius =\sqrt{(-\cos \theta)^{2}+(-\sin \theta)^{2}-(-8)}

=\sqrt{1+8}

= 3

 

(iv) x^{2}+y^{2}-a x-b y=0

Sol :

The given equation can be rewritten as x^{2}+y^{2}-\frac{2 a x}{2}-\frac{2 b y}{2}=0

Center =\left(\frac{a}{2}, \frac{b}{2}\right)

radius =\sqrt{\left(\frac{a}{2}\right)^{2}+\left(\frac{b}{2}\right)^{2}}

=\frac{1}{2} \sqrt{a^{2}+b^{2}}

 

QUESTION 2

Find the equation of the circle passing through the points:

(i) (5,7),(8,1) and (1,3)

Sol :

Let the required circle be x^{2}+y^{2}+2 g x+2 f y+c=0 . \quad \ldots .(1)

It passes through (5,7),(8,1) and (1,3)

Substituting the coordinates of these points in equation (1) :

74+10 g+14 f+c=0 \quad \ldots(2)

65+16 g+2 f+c=0 \quad \ldots(3)

10+2 g+6 f+c=0 \quad \ldots .(4)

Simplifying (2),(3) and (4)

g=\frac{-29}{\phantom{-}6} f=\frac{-19}{\phantom{-}6} c=\frac{56}{3}

Equation of the required circle:

x^{2}+y^{2}-\frac{29 x}{3}-\frac{19 y}{3}+\frac{56}{3}=0

\Rightarrow 3\left(x^{2}+y^{2}\right)-29 x-19 y+56=0

 

(ii) (1,2),(3,-4) and (5,-6)

Sol :

Let the required circle be x^{2}+y^{2}+2 g x+2 f y+c=0 . \ldots(1)

It passes through (1,2),(3,-4) and (5,-6)

Substituting the coordinates of these points in equation (1) :

5+2 g+4 f+c=0

25+6 g-8 f+c=0 \quad \ldots(3)

61+10 g-12 f+c=0 \quad \ldots(4)

Simplifying (2),(3) and (4)

g=-11  f=-2 c=25

The equation of the required circle is x^{2}+y^{2}-22 x-4 y+25=0

 

(iii) (5,-8),(-2,9) and (2,1)

Sol :

Let the required circle be x^{2}+y^{2}+2 g x+2 f y+c=0 . \ldots (1)

It passes through (5,-8),(-2,9) and (2,1)

Substituting the coordinates of these points in equation (1):

89+10 g-16 f+c=0 \quad \ldots(2)

85-4 g+18 f+c=0 \quad \ldots (3)

5+4 g+2 f+c=0 \quad \ldots(4)

Simplifying (2),(3) and (4)

g=58 f=24 c=-285

The equation of the required circle is x^{2}+y^{2}+116 x+48 y-285=0

 

(iv) (0,0),(-2,1) and (-3,2)

Sol :

Let the required circle be x^{2}+y^{2}+2 g x+2 f y+c=0 . \ldots(1)

It passes through (0,0),(-2,1) and (-3,2)

Substituting the coordinates of these points in equation (1):

c=0\dots(2)

5-4 g+2 f+c=0 \quad \ldots(3)

13-6 g+4 f+c=0 \quad \ldots(4)

Simplifying (2),(3) and (4)

g=\frac{-3}{2} f=\frac{-11}{2} c=0

The equation of the required circle is x^{2}+y^{2}-3 x-11 y=0

 

QUESTION 3

Find the equation of the circle which passes through (3,-2),(-2,0) and has its center on the line
2x-y=3 .

Sol 

Let the required equation of the circle be x^{2}+y^{2}+2 g x+2 f y+c=0 . \ldots (1)

It is given that the circle passes through (3,-2),(-2,0)

13+6 g-4 f+c=0 \ldots \ldots(2)

4-4 g+c=0

The center lies on the line 2 x-y=3

-2 g+f-3=0 \quad \ldots(4)

Solving (2),(3) and (4) : g=\frac{3}{2}f=6c=2 Hence, the required equation of circle isx^{2}+y^{2}+3 x+12 y+2=0

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<span style="background-color: #ffff00;">QUESTION 4</span>
Find the equation of the circle which passes through the points

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(3,7),(5,5)and has its center on the linex-4 y=1. Sol : Let the required equation of the circle bex^{2}+y^{2}+2 g x+2 f y+c=0 . \ldots(1) It is given that the circle passes through(3,7),(5,5) 58+6 g+14 f+c=0 \ldots(2) 50+10 g+10 f+c=0 \quad \ldots(3)

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The center

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(-g,-f)lies on the linex-4 y=1 -g+4 f-1=0 \quad \ldots .(4) Solving (2),(3) and(4) :

g=3 f=1 c=-90

Hence, the required equation of the circle is x^{2}+y^{2}+6 x+2 y-90=0

 

QUESTION 5

Show that the points (3,-2),(1,0),(-1,-2) and (1,-4) are concyclic.

Sol :

Let the required equation of the circle be x^{2}+y^{2}+2 g x+2 f y+c=0 . \quad \ldots(1)

It is given that the circle passes through (3,-2),(1,0),(-1,-2)

13+6 g-4 f+c=0 \quad \ldots(2)

1+2 g+c=0 \quad \ldots(3)

5-2 g-4 f+c=0 \quad \ldots(4)

Solving (2),(3) and (4)

g=-1 f=2 c=1

Therefore, the equation of the circle is x^{2}+y^{2}-2 x+4 y+1=0 . \ldots(5)

We see that the point (1,-4) satisfies the equation (5)

Hence, the points (3,-2),(1,0),(-1,-2) and (1,-4) are concyclic.

 

QUESTION 6

Show that the points (5,5),(6,4),(-2,4) and (7,1) all lie on a circle, and find its equation, center and radius.

Sol :

Let the required equation of the circle be x^{2}+y^{2}+2 g x+2 f y+c=0 . \quad \ldots(1)

It is given that the circle passes through (5,5),(6,4),(-2,4)

50+10 g+10 f+c=0 \quad \ldots(2)

52+12 g+8 f+c=0 \quad \ldots(3)

20-4 g+8 f+c=0 \quad \ldots(4)

Solving (3) and (4) by subtracting ,we get

g=-2

then substituting value of in equation (2) and (3) we get

30+10f+c=0\dots(5)

28+8f+c=0\dots(6)

Solving (5) and (6) by subtracting , we get

f=-1

then substituting value of f in equation (6) , we get 

c=-20

Thus, the equation of the circle is x^{2}+y^{2}-4 x-2 y-20=0 . \ldots(7)

We see that the point (7,1) satisfies equation (7)

Hence, the points (5,5),(6,4),(-2,4) and (7,1) lie on the circle.

Also, center of the required circle =(2,1)

Radius of the required circle =\sqrt{4+1+20}=5

 

QUESTION 7

Find the equation of the circle which circumscribes the triangle formed by the lines :

(i) x+y+3=0, x-y+1=0 and x=3

Sol :

In \triangle A B C :

Let A B represent the line x+y+3=0

Let BC represent the line x-y+1=0

Let CA represent the line x=3

 

Let A,B,C are the points of intersection of lines (i) and (ii) , (ii) and (iii) , (iii) and (i) respectively

A=(-2,-1) B=(3,4) C=(3,-6) are the co-ordinates

Now A circle x^{2}+y^{2}+2 g x+2 f y+c=0

Circle passes through (-2,-1) (3,4) (3,-6)

45+6 g-12 f+c=0\dots(1)

5-4 q-2 f+c=0\dots(2)

25+6 g+8 f+c=0\dots(3)

Solving (1),(2),(3) we get

g=-3 f=1 c=-15

Hence, the required equation of the circumcircle is x^{2}+y^{2}-6 x+2 y-15=0

 

(ii) 2 x+y-3=0, x+y-1=0 and 3 x+2 y-5=0

Sol :

In \triangle \mathrm{ABC} :

Let AB represent the line 2 x+y-3=0 . \ldots (1) 

Let BC represent the line x+y-1=0 . \quad \ldots(2)

Let CA represent the line 3 x+2 y-5=0 . . . .(3)

 

Intersection point of (1) and (3) is (1,1)

Intersection point of (1) and (2) is (2,-1)

Intersection point of (2) and (3) is (3,-2)

 

The coordinates of A, B and C are (1,1),(2,-1) and (3,-2), respectively.

Let the equation of the circumcircle be x^{2}+y^{2}+2 g x+2 f y+c=0
It passes through \mathrm{A}, \mathrm{B} and \mathrm{C} .

2+2 g+2 f+c=0

5+4 g-2 f+c=0

13+6 g-4 f+c=0

g=\frac{-13}{2} f=\frac{-5}{2} c=16

Hence, the required equation of the circumcircle is x^{2}+y^{2}-13 x-5 y+16=0

 

(iii) x+y=2,3 x-4 y=6 and x-y=0

Sol :

In \triangle \mathrm{ABC} :

Let AB represent the line

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Let BC represent the line 3 x-4 y=6 . \quad \ldots(2)

Let CA represent the line x-y=0 . \quad \ldots (3)

 

Intersection point of (1) and (3) is (1,1)

Intersection point of (1) and (2) is (2,0)

Intersection point of (2) and (3) is (-6,-6)

 

The coordinates of \mathrm{A}, \mathrm{B} and \mathrm{C} are (1,1),(2,0) and (-6,-6), respectively.

Let the equation of the circumcircle be x^{2}+y^{2}+2 g x+2 f y+c=0

It passes through A,B and C

2+2 g+2 f+c=0

4+4 g+c=0

72-12 g-12 f+c=0

g=2 f=3 c=-12

Hence, the required equation of the circumcircle is x^{2}+y^{2}+4 x+6 y-12=0

 

(iv) y=x+2,3 y=4 x and 2 y=3 x

Sol :

In \triangle A B C :

Let A B represent the line y=x+2 \quad \ldots (1) 

Let BC represent the line 3 y=4 x \quad \ldots (2)

Let CA represent the line 2 y=3 x \quad \ldots (3)

 

Intersection point of (1) and (3) is (4,6)

Intersection point of (1) and (2) is (6,8)

Intersection point of (2) and (3) is (0,0)

 

Therefore, the coordinates of \mathrm{A}, \mathrm{B} and \mathrm{C} are (4,6),(6,8) and (0,0) respectively.

 

Let the equation of the circumcircle be x^{2}+y^{2}+2 g x+2 f y+c=0
It passes through \mathrm{A}, \mathrm{B} and \mathrm{C} .

52+8 g+12 f+c=0

100+12 g+16 f+c=0

c=0

g=-23 f=11 c=0

Hence, the required equation of the circumcircle is x^{2}+y^{2}-46 x+22 y=0

 

QUESTION 8

Prove that the centers of the three circles x^{2}+y^{2}-4 x-6 y-12=0, x^{2}+y^{2}+2 x+4 y-10=0 and x^{2}+y^{2}-10 x-16 y-1=0 are collinear.

Sol :

The given equations of the circles are as follows:

x^{2}+y^{2}-4 x-6 y-12=0, \ldots(1)

x^{2}+y^{2}+2 x+4 y-10=0 \quad \ldots(2)

x^{2}+y^{2}-10 x-16 y-1=0 \quad \ldots(3)

 

The center of circle (1) is (2,3)

The center of circle (2) is (-1,-2)

The center of circle (3) is (5,8)

 

The area of the triangle formed by the points (2,3),(-1,-2) and (5,8) is

=\frac{1}{2}|2(-10)-1(5)+5(5)|

=\frac{1}{2}|-25+25|

= 0

Hence, the centers of the circles x^{2}+y^{2}-4 x-6 y-12=0, x^{2}+y^{2}+2 x+4 y-10=0 and x^{2}+y^{2}-10 x-16 y-1=0 are collinear.

 

QUESTION 9

Prove that the radii of the circles x^{2}+y^{2}=1, x^{2}+y^{2}-2 x-6 y-6=0 and x^{2}+y^{2}-4 x-12 y-9=0 are in A.P.

Sol :

Let the radii of the circles x^{2}+y^{2}=1 , x^{2}+y^{2}-2 x-6 y-6=0 and x^{2}+y^{2}-4 x-12 y-9=0 be r_{1}, r_{2} and r_{3, \text { respectively. }}

r_{1}=1 , r_{2}=\sqrt{(-1)^{2}+(-3)^{2}+6}=4 , r_{3}=\sqrt{(-2)^{2}+(-6)^{2}+9}=7

Now , r_{2}-r_{1}=r_{3}-r_{2}=3

r_{1}, r_{2} and r_{3} are in A.P

 

QUESTION 10

Find the equation of the circle which passes through the origin and cuts off intercept of lengths 4
and 6 on the positive side of the x-axis and y-axis respectively.

Sol :

According to the question, the circle passes through the origin.

Let the equation of the circle be x^{2}+y^{2}-2 h x-2 k y=0

The circle cuts off chords of lengths 4 and 6 on the positive sides of the x-axis and the y- axis, respectively.

Center =\left(\frac{4}{2}, \frac{6}{2}\right)

=(2,3)=(h, k)

Required equation :

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

\Rightarrow x^{2}+y^{2}-4 x-6 y=0

 

QUESTION 11

Find the equation of the circle concentric with the circle x^{2}+y^{2}-6 x+12 y+15=0 and double of its area.

Sol :

Let the equation of the required circle be x^{2}+y^{2}+2 g x+2 f y+c=0

The center of the circle x^{2}+y^{2}-6 x+12 y+15=0 is (3,-6)

Area of the required circle =2 \pi r^{2}

Here, r= radius of the given circle

r=\sqrt{9+36-15}

=\sqrt{30}

Area of the required circle =2 \pi(30)=60 \pi

Let R be the radius of the required circle.

60 \pi=\pi R^{2}

R^{2}=60

Thus, the equation of the required circle is (x-3)^{2}+(y+6)^{2}=60, i.e. x^{2}+y^{2}-6 x+12 y=15

 

QUESTION 12

Find the equation to the circle which passes through the points (1,1)(2,2) and whose radius is 1 .Show that there are two such circles.

Sol :

Let the equation of the required circle be x^{2}+y^{2}+2 g x+2 f y+c=0

It passes through (1,1) and (2,2)

2 g+2 f+c=-2 \ldots .(1)

4 g+4 f+c=-8 \ldots .(2)

From (1) and (2), we have:

-2 g-2 f=6

g+f=-3 \quad \ldots(3)

From (2) and (3), we have:

c=4

 

\Rightarrow \sqrt{g^{2}+f^{2}-c}=1

\Rightarrow g^{2}+f^{2}=1+c=5

\Rightarrow(g+f)^{2}-2 g f=5

\Rightarrow g f=2 \dots (4)

 

by using (g-f)^2=(g+f)^2-4gf

g-f=\pm 1 \dots(5)

 

Solving (4) and (5) , we get

g=-1 \text{ or } -2

f=-2 \text{ or } -1

Therefore, the required equations of the circles are x^{2}+y^{2}-4 x-2 y+4=0 and x^{2}+y^{2}-2 x-4 y+4=0

Hence, there are two such circles.

 

QUESTION 13

Find the equation of the circle concentric with x^{2}+y^{2}-4 x-6 y-3=0 and which touches the y-axis.

Sol :

since, the circles are concentric \Rightarrow Centre of required circle = Centre of x^{2}+y^{2}-4 x-6 y-3=0

The centre of the required circle is (2,3)

We know that if a circle with centre (h, k) touches the y -axis, then h is the radius of the circle.
Thus, the radius is 2 

Equation of the circle:

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

\Rightarrow x^{2}+y^{2}-4 x-6 y+9=0

 

QUESTION 14

If a circle passes through the point (0,0) (a, 0) (0, b) (0, b) then find the coordinates of its center .

Sol :

The general equation of the circle is x^{2}+y^{2}+2 g x+2 f y+c=0

Now, it is passing through (0,0)

\therefore c=0

Also, it is passing through (a, 0)

\therefore a^{2}+2 a g=0

\Rightarrow a(a+2 g)=0

\Rightarrow a+2 g=0

\Rightarrow g=-\frac{a}{2}

Again, it is passing through (0, b)

\therefore b^{2}+2 b f=0

\Rightarrow b(b+2 f)=0

\Rightarrow b+2 f=0

\Rightarrow f=-\frac{b}{2}

The coordinates of its centre are given by

(-g,-f)=\left(\frac{a}{2}, \frac{b}{2}\right)

 

QUESTION 15

Find the equation of the circle which passes through the points (2,3) and (4,5) and the center lies on the straight line y-4 x+3=0

Sol :

The general equation of the circle is x^{2}+y^{2}+2 g x+2 f y+c=0 where the center of the circle is (-g,-f )

Now, it is passing through (2,3)

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

Also, it is passing through (4,5)

41+8 g+10 f+c=0 \dots (2)

Now, the center lies on the straight line y-4 x+3=0  , (-g , -f) satisfies this equation 

-f+4 g+3=0 \quad \ldots \ldots(3)

Solving (1),(2) and (3), we get

g=-2, f=-5 and c=25

The equation of the circle is given by x^{2}+y^{2}-4 x-10 y+25=0

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