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 $Hence, the required equation of circle is$ x^{2}+y^{2}+3 x+12 y+2=0

*** QuickLaTeX cannot compile formula:

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

*** Error message:
Missing $ inserted.

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

*** QuickLaTeX cannot compile formula:

 
The center

*** Error message:
Missing $ inserted.

(-g,-f) $lies on the line$ x-4 y=1 $\text{[math]}$ -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 g 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

*** QuickLaTeX cannot compile formula:
x+y=2 .\quad \Idots(1)

*** Error message:
Undefined control sequence \Idots.
leading text: $x+y=2 .\quad \Idots

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 $x^{2}+y^{2}-4 x-10 y+25=0$

myhelper

A quality solution

Circle

EXERCISE 24.2

Leave a Reply Cancel reply