EXERCISE 24.1

QUESTION 1

Find the equation of the circle with :

(i) Center $(-2,3)$ and radius 4

Sol :

Let $(h, k)$ be the centre of a circle with radius a.

Thus, its equation will be $(x-h)^{2}+(y-k)^{2}=a^{2}$

Here, $h=-2, k=3$ and $a=4$

$\therefore$ Required equation of the circle:

$\Rightarrow (x+2)^{2}+(y-3)^{2}=4^{2}$

$\Rightarrow(x+2)^{2}+(y-3)^{2}=16$

 

(ii) Centre $(a, b)$ and radius $\sqrt{a^{2}+b^{2}}$

Sol :

Here, $h=a, k=b$ and radius $=\sqrt{a^{2}+b^{2}}$

$\therefore$ Required equation of the circle:

$\Rightarrow (x-a)^{2}+(y-b)^{2}=a^{2}+b^{2}$

$\Rightarrow x^{2}+y^{2}-2 a x-2 b y=0$

 

(iii) Centre $(0,-1)$ and radius 1

Sol :

Here, $h=0, k=-1$ and radius $=1$

$\therefore$ Required equation of the circle:

$(x-0)^{2}+(y+1)^{2}=(1)^{2}$

$\Rightarrow x^{2}+y^{2}+2 y=0$

 

(iv) Centre $(a \cos a, a \sin \alpha)$ and radius a.

Sol :

Here, $h=a \cos \alpha, k=a \sin \alpha$ and radius $=a$

$\therefore$ Required equation of the circle:

$(x-a \cos \alpha)^{2}+(y-a \sin \alpha)^{2}=(a)^{2}$

$\Rightarrow x^{2}+a^{2} \cos ^{2} \alpha-2 a x \cos \alpha+y^{2}+a^{2} \sin ^{2} \alpha-2 a y \sin \alpha=a^{2}$

$\Rightarrow x^{2}+a^{2}\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)-2 a x \cos \alpha+y^{2}-2 a y \sin \alpha=a^{2}$

$\Rightarrow x^{2}+a^{2}-2 a x \cos \alpha+y^{2}-2 a y \sin \alpha=a^{2}$

$\Rightarrow x^{2}+y^{2}-2 a x \cos \alpha-2 a y \sin \alpha=0$

 

(v) Centre $(a, a)$ and radius $\sqrt{2} a$

Sol :

Here, $h=a, k=a$ and radius $=\sqrt{2} a$

$\therefore$ Required equation of the circle:

$(x-a)^{2}+(y-a)^{2}=(\sqrt{2} a)^{2}$

$\Rightarrow x^{2}+a^{2}-2 a x+y^{2}+a^{2}-2 a y=2 a^{2}$

$\Rightarrow x^{2}+y^{2}-2 a y-2 a x=0$

 

QUESTION 2

Find the center and radius of each of the following circles:

(i) $(i)(x-1)^{2}+y^{2}=4$

Sol :

Let $(h, k)$ be the centre of a circle with radius a.

Thus, its equation will be $(x-h)^{2}+(y-k)^{2}=a^{2}$

 

Given: $(x-1)^{2}+y^{2}=4$

Here, $h=1, k=0$ and $a=2$

Thus, the center is $(1,0)$ and the radius is 2

 

(ii) (ii) $(x+5)^{2}+(y+1)^{2}=9$

Sol :

Given: $(x+5)^{2}+(y+1)^{2}=9$

Here, $h=-5, k=-1$ and radius $=3$

Thus, the center is $(-5,-1)$ and the radius is 3

 

(iii) $x^{2}+y^{2}-4 x+6 y=5$

Sol :

Given: $x^{2}+y^{2}-4 x+6 y=5$

The given equation can be rewritten as follows:

$(x-2)^{2}+(y+3)^{2}-4-9=5$

$\Rightarrow(x-2)^{2}+(y+3)^{2}=18$

Thus, the centre is $(2,-3)$

And, radius $=\sqrt{18}=3 \sqrt{2}$

 

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

Sol :

Given:  $x^{2}+y^{2}-x+2 y-3=0$

The given equation can be rewritten as follows:

$\left(x-\frac{1}{2}\right)^{2}+(y+1)^{2}-\frac{1}{4}-1-3=0$

$\Rightarrow\left(x-\frac{1}{2}\right)^{2}+(y+1)^{2}=\frac{17}{4}$

Thus, the center is $\left(\frac{1}{2},-1\right)$ and and the radius is $\frac{\sqrt{17}}{2}$

 

QUESTION 3

Find the equation of the circle whose center is $(1,2)$ and which passes through the point $(4,6)$

Sol :

Let $(h, k)$ be the center of a circle with radius a.

Thus, its equation will be $(x-h)^{2}+(y-k)^{2}=a^{2}$

Given: $h=1, k=2$

Equation of the circle $=(x-1)^{2}+(y-2)^{2}=a^{2} \quad \ldots(1)$

Also, equation $(1)$ passes through $(4,6)$

$\Rightarrow (4-1)^{2}+(6-2)^{2}=a^{2}$

$(4-1)^{2}+(6-2)^{2}=a^{2}$

$\Rightarrow 9+16=a^{2}$

$\Rightarrow a=5$ $(\because a>0)$

 

Substituting the value of a in equation $(1) :$

$\Rightarrow (x-1)^{2}+(y-2)^{2}=25$

$\Rightarrow x^{2}+1-2 x+y^{2}+4-4 y=25$

$\Rightarrow x^{2}-2 x+y^{2}-4 y=20$

$\Rightarrow x^{2}+y^{2}-2 x-4 y-20=0$

Thus, the required equation of the circle is $x^{2}+y^{2}-2 x-4 y-20=0$

 

QUESTION 4

Find the equation of the circle passing through the point of intersection of the lines $x+3 y=0$ and $2 x-7 y=0$ and whose center is the point of intersection of the lines $x+y+1=0$ and $x-2 y+4=0$

Sol :

The point of intersection of the line $x+3y=0$ and $2x-7y=0$ can be find by simultaneously solving the equations , that is $(0,0)$

Let $(h,k)$ be the center of a circle with radius a 

Thus , its equation will be $(x-h)^{2}+(y-k)^{2}=a^{2}$

The point of intersection of the lines $x+y+1=0$ and $x-2 y+4=0$ is can be find by simultaneously solving the equation and points are $(-2,1)$

$\therefore h=-2, k=1$

Equation of the required circle $=(x+2)^{2}+(y-1)^{2}=a^{2} \quad \ldots(1)$

Also, equation (1) passes through $(0,0)$

$\Rightarrow (0+2)^{2}+(0-1)^{2}=a^{2}$

$\Rightarrow 4+1=a^{2}$

$\Rightarrow a=\sqrt{5}$ $(\because a>0)$

 

Substituting the value of a in equation (1) :

$\Rightarrow (x+2)^{2}+(y-1)^{2}=5$

$\Rightarrow x^{2}+4+4 x+y^{2}+1-2 y=5$

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

Hence, the required equation of the circle is $x^{2}+y^{2}+4 x-2 y=0$

 

QUESTION 5

Find the equation of the circle whose center lies on the position of y – axis at a distance 6 from the origin and whose radius is 4 

Sol :

Let $(h, k)$ be the center of a circle with radius a

Thus, its equation will be $(x-h)^{2}+(y-k)^{2}=a^{2}$

The center of the required circle lies on the positive direction of the y-axis at a distance 6 from the origin 

Thus , the co-ordinates of the center are $(0,6)$

$h=0, k=6$

Equation of the circle $=(x-0)^{2}+(y-6)^{2}=a^{2} \quad \ldots(1)$

Also, $a=4$

Substituting the value of a in equation (1) :

$\Rightarrow (x-0)^{2}+(y-6)^{2}=16$

$\Rightarrow x^{2}+y^{2}+36-12 y=16$

$\Rightarrow x^{2}+y^{2}-12 y+20=0$

Hence, the required equation of the circle is $x^{2}+y^{2}-12 y+20=0$

 

QUESTION 6

If the equations of two diameters of a circle are $2 x+y=6$ and $3 x+2 y=4$ and the radius is 10 find the equation of the circle 

Sol :

Let $(h, k)$ be the center of a circle with radius a

Thus, its equation will be $(x-h)^{2}+(y-k)^{2}=a^{2}$

The intersection point of $2 x+y=6$ and $3 x+2 y=4$ is $(8,-10)$ find by simultaneously solving given equation 

The diameters of a circle intersect at the center

Thus, the coordinates of the center are $(8,-10)$

$\therefore h=8, k=-10$

Thus, the equation of the required circle is $(x-8)^{2}+(y+10)^{2}=a^{2} \quad \ldots(1)$

Also , $a=10$

Substituting the value of $a$ in equation $(1) :$

$\Rightarrow (x-8)^{2}+(y+10)^{2}=100$

$\Rightarrow x^{2}+y^{2}-16 x+64+100+20 y=100$

$\Rightarrow x^{2}+y^{2}-16 x+20 y+64=0$

Hence, the required equation of the circle is $x^{2}+y^{2}-16 x+20 y+64=0$

 

QUESTION 7

Find the equation of a circle

(i) Which touches both the axes at a distance of 6 units from the origin.

Sol :

Let $(h, k)$ be the center of a circle with radius a

Thus, its equation will be $(x-h)^{2}+(y-k)^{2}=a^{2}$

Let the required equation of the circle be $(x-h)^{2}+(y-k)^{2}=a^{2}$

It is given that the circle passes through the points $(6,0)$ and $(0,6)$

$(6-h)^{2}+(0-k)^{2}=6^{2}$

$\Rightarrow(6-h)^{2}+(-k)^{2}=36$

$\Rightarrow 36+h^{2}-12 h+k^{2}=36$

$\Rightarrow h^{2}+k^{2}=12 h \quad \ldots(1)$

AND

$(0-h)^{2}+(6-k)^{2}=6^{2}$

$\Rightarrow h^{2}+36+k^{2}-12 k=36$

$\Rightarrow h^{2}+k^{2}=12 k \quad \ldots(2)$

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

$12 k=12 h \Rightarrow h=k$

 

$\therefore$ From equation (2), we have:

$\Rightarrow k^{2}+k^{2}=12 k$

$\Rightarrow k^{2}-6 k=0$

$\Rightarrow k(k-6)=0$

$\Rightarrow k=6$ $(\because k>0)$

 

Consequently, we get: $h=6$

Hence, the required equation of the circle is

$(x-6)^{2}+(y-6)^{2}=36$ or $x^{2}+y^{2}-12 x-12 y+36=0$

 

(ii) Which touches x-axis at a distance 5 from the origin and radius 6 units.

Sol :

The circle touches the $x-$ axis at $A=(5,0)$ and has radius 6 unit

Thus , center $=(5, b)$

$\therefore h=5,k=b$ $a=6$

Let the required equation of the circle be $(x-h)^{2}+(y-k)^{2}=a^{2}$

$\Rightarrow (5-5)^2+(0-b)^2=6^2$

$\Rightarrow b=6$

$\Rightarrow$ centre $=(5,6)$

so, the equation of required circle is

$(x-5)^{2}+(y-6)^{2}=6^{2}$

$\Rightarrow \quad x^{2}+y^{2}-10 x-12 y+25=0$

 

(iii) Which touches both the axes and passes through the point $(2,1)$ .

Sol :

Let the required equation of the circle be $(x-h)^{2}+(y-k)^{2}=a^{2}$

It is given that the circle touches both the axes.

Thus, the required equation will be $x^{2}+y^{2}-2 a x-2 a y+a^{2}=0$

Also, the circle passes through the point $(2,1)$

$\Rightarrow 4+1-4 a-2 a+a^{2}=0$

$\Rightarrow a^{2}-6 a+5=0$

$\Rightarrow a^{2}-5 a-a+5=0$

$\Rightarrow a=1,5$

Hence, the required equation is $x^{2}+y^{2}-2 x-2 y+1=0$ or $x^{2}+y^{2}-10 x-10 y+25=0$

 

(iv) Passing through the origin, radius 17 and ordinate of the center is $-15 $

Sol :

Let the required equation of the circle be $(x-h)^{2}+(y-k)^{2}=a^{2}$

Given: $k=-15, a=17$

Also , circle passes through the point $(0,0)$

$\therefore$ Equation of the circle:

$(0-h)^{2}+(0-15)^{2}=(17)^{2}$

$\Rightarrow h=\pm 8$

Hence, the required equation of the circle is $(x-8)^{2}+(y+15)^{2}=17^{2}$ or $(x+8)^{2}+(y+15)^{2}=17^{2},$ i.e. $x^{2}+y^{2} \pm 16 x+30 y=0$

 

QUESTION 8

Find the equation of the circle which has its center at the point $(3,4)$ and touches the straight line $5 x+12 y-1=0$

Sol :

It is given that the center is at the point $(3,4)$

Let the equation of the circle be $(x-h)^{2}+(y-k)^{2}=a^{2}$

Equation of the required circle $=(x-3)^{2}+(y-4)^{2}=a^{2} \ldots \ldots(1)$

 

Also, the circle touches the straight line $5 x+12 y-1=0$ to which points $(3,4)$ satisfies the equation $[\because$ radius is perpendicular to the tangent $]$

$\therefore a=\left|\frac{5(3)+12(4)-1}{\sqrt{5^{2}+12^{2}}}\right|$ $=\left|\frac{62}{13}\right|$

$\Rightarrow a^{2}=\left|\frac{5(3)+12(4)-1}{13}\right|$ $=\frac{3844}{169}$ $=\frac{62}{13}$

So, from equation (1) ,we have:

$(x-3)^{2}+(y-4)^{2}=\left(\frac{62}{13}\right)^{2}$

$\Rightarrow \quad 169\left[x^{2}+y^{2}-6 x-8 y\right]+25 \times 169=3844$

$\Rightarrow \quad 169\left[x^{2}+y^{2}-6 x-8 y\right]+381=0$

Hence, the required equation of the circle is $169\left(x^{2}+y^{2}-6 x-8 y\right)+381=0$

 

QUESTION 9

Find the equation of the circle which touches the axes and whose center lies on $x-2 y=3$

Sol :

The required circle touches $A(a, 0)$ and $B(0, a)$ on the axes

so the center $=(a, a)  $ radius $=a$

Also, the center lies on $x-2 y=3$

$\Rightarrow \quad a-2 a=3$

$\Rightarrow \quad-a=3$

$\Rightarrow a=-3$

$\therefore$ center $=(-3,-3)$ (this shows center in third quadrant) and radius $=3$

Thus the equation of circle is

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

$\Rightarrow \quad x^{2}+y^{2}+6 x+6 y+9=0$

 

If the circle lies in the fourth quadrant, then its center will be $\left(a_{,}-a\right)$

$\therefore a+2 a=3 $

$\Rightarrow a=1$

$\therefore$ Required equation of the circle $=(x-1)^{2}+(y+1)^{2}=1$

$=x^{2}+y^{2}-2 x+2 y+1=0$

 

QUESTION 10

A circle whose center is the point of intersection of the lines $2 x-3 y+4=0$ and $3 x+4 y-5=0$
passes through the origin. Find its equation.

Sol :

Let the required equation of the circle be $(x-h)^{2}+(y-k)^{2}=a^{2}$

We have ,

$2 x-3 y=-4 \ldots \ldots \ldots(1)$

$3 x+4 y=5 . . . . . . . .(2)$

The point of intersection of (1) and (2) is

$=\left(\frac{-1}{17}, \frac{22}{17}\right)$

According to the equation center $=\left(\frac{-1}{17}, \frac{22}{17}\right)$

Also, circle passes through origin $(0,0)$

$\left(0-\dfrac{1}{17}\right)^2 + \left(0-\dfrac{22}{17}\right)^2=a^2$

$\frac{1}{289}+\frac{484}{289}=a^2$

$a^2=\frac{485}{289}$

Hence, the required equation of the circle is $\left(x+\frac{1}{17}\right)^{2}+\left(y-\frac{22}{17}\right)^{2}=\frac{485}{288}$

 

QUESTION 11

A circle of radius 4 units touches the coordinate axes in the first quadrant. Find the equations of
its images with respect to the line mirrors $x=0$ and $y=0$ 

Sol :