# EXERCISE 24.3

QUESTION 1

Find the equation of the circle, the end points of whose diameter are $(2,-3)$ and $(-2,4)$ . Find its center and radius.

Sol :

$(2,-3)$ and $(-2,4)$ are the ends points of the diameter of a circle. The equation of this circle is $(x-2)(x+2)+(y+3)(y-4)=0$

$\Rightarrow x^{2}-4+y^{2}-4 y+3 y-12=0$

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

Equation (1) can be rewritten as

$\Rightarrow x^{2}+\left(y-\frac{1}{2}\right)^{2}-\frac{1}{4}-16=0$

$\Rightarrow x^{2}+\left(y-\frac{1}{2}\right)^{2}=\frac{65}{4}$

$\therefore$ Center is $\left(0, \frac{1}{2}\right)$ and radius is $\frac{\sqrt{65}}{2}$

QUESTION 2

Find the equation of the circle the end points of whose diameter are the centers of the circles $x^{2}+y^{2}+6 x-14 y-1=0$ and $x^{2}+y^{2}-4 x+10 y-2=0$

Sol :

Given :

$x^{2}+y^{2}+6 x-14 y-1=0 \ldots(1)$

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

Equations $(1)$ and $(2)$ can be rewritten as follows:

$(x+3)^{2}+(y-7)^{2}=59$

$(x-2)^{2}+(y+5)^{2}=31$

Thus, the centers of the circles are $(-3,7)$ and $(2,-5)$

Hence, the equation of the circle, the end points of whose diameter are the centers of the given
circles, is $(x+3)(x-2)+(y-7)(y+5)=0,$ i.e. $x^{2}+y^{2}+x-2 y-41=0$

QUESTION 3

The sides of a square are $x=6, x=9, y=3$ and $y=6 .$ Find the equation of a circle drawn on the diagonal of the square as its diameter.

Sol :

According to the question:

Sides of the square are $x=6$ $x=9$ $y=3$ and $y=6$

The vertices of the square are $(6,6)$ $(9,6)$ $(9,3)$ and $(6,3) .$

And, the vertices of two diagonals are $(6,6)$ $(9,3)$ and $(9,6)$ $(6,3)$

Hence, the equation of the circle is $(x-6)(x-9)+(y-6)(y-3)$ or $x^{2}+y^{2}-15 x-9 y+72=0$

QUESTION 4

Find the equation of the circle circumscribing the rectangle whose sides are $x-3 y=4$ $3x+y=22$ $x-3 y=14$ and $3 x+y=62$

Sol :

Sides of the rectangle:

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

$3 x+y=22 \quad \ldots(2)$

$x-3 y=14 \quad \ldots(3)$

$3 x+y=62 \quad \ldots(4)$

The intersection of (1) and (2) is $(7,1)$

The intersection of (2) and (3) is $(8,-2)$

The intersection of (3) and (4) is $(20,2)$ .

The intersection of (1) and (4) is $(19,5)$ .

Hence, the vertices of the rectangle are $(7,-1),(8,-2),(20,2)$ and $(19,5)$

The vertices of the diagonals are $(7,-1),(20,2)$ and $(19,5),(8,-2)$

Thus, the required equation of the circle is $(x-7)(x-20)+(y-1)(y-2)=0$ or $x^{2}+y^{2}-27 x-3 y+142=0$

QUESTION 5

Find the equation of the circle passing through the origin and the points where the line $3x+4y=12$ meets the axes of coordinates.

Sol :

Putting $x=0$ in $3 x+4 y=12$ :

$y=3$

Putting $y=0$ in $3 x+4 y=12 :$

$x=4$

Thus, the line $3 x+4 y=12$ meets the axes of coordinates at points $A(0,3)$ and $B(4,0)$

The equation of the circle with AB as the diameter is $(x-0)(x-4)+(y-3)(y-0)=0$ or $x^{2}-4 x+y^{2}-3 y=0$

Hence, the required equation is $x^{2}-4 x+y^{2}-3 y=0$

QUESTION 6

Find the equation of the circle which passes through the origin and cuts off intercepts a and b
respectively from x and y-axes.

Sol :

CASE-1

If the required circle passes through the origin and $(a, b),$ then the end points of the diameter of the circle will be $(0,0)$ and $(a, b)$ and $(a, b)$

Required equation of circle:

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

CASE-2

If the required circle passes through the origin and $(-a,-b),$ then the end points of the diameter of the circle will be $(0,0)$ and $(-a,-b)$

Required equation of circle :

$(x-0)(x+a)+(y-0)(y+b)$

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

Hence, the equation of the required circle is $x^{2}+y^{2} \pm a x \pm b y=0$

QUESTION 7

Find the equation of the circle whose diameter is the line segment joining $(-4,3)$ and $(12,-1)$ Find also the intercept made by it on y-axis.

Sol :

It is given that the end points of the diameter of the circle are $(-4,3)$ and $(12,-1)$

Required equation of circle :

$(x+4)(x-12)+(y-3)(y+1)$
$x^{2}+y^{2}-8 x-2 y-51=0 \quad \ldots(1)$

Putting $x=0$ in (1) :

$y^{2}-2 y-51=0$

$\Rightarrow y^{2}-2 y-51=0$

$\Rightarrow y=1 \pm 2 \sqrt{13}$

Hence, the intercepts made by it on the $y$ -axis is $1+2 \sqrt{13}-1+2 \sqrt{13}=4 \sqrt{13}$

QUESTION 8

The abscissae of the two points A and B are the roots of the equation $x^{2}+2 a x-b^{2}=0$ and their ordinates are the roots of the equation $x^{2}+2 p x-q^{2}=0$ Find the equation of the circle with AB as diameter. Also, find its radius.

Sol :

Roots of equation $x^{2}+2 a x-b^{2}=0$ are

$x=\frac{-2 a \pm \sqrt{4 a^{2}+4 b^{2}}}{2}$ $=-a \pm \sqrt{a^{2}+b^{2}}$

Roots of equation $x^{2}+2 p x-q^{2}=0$ are

$y=\frac{-2 p \pm \sqrt{4 p^{2}+4 q^{2}}}{2}$ $=-P \pm \sqrt{p^{2}+q^{2}}$

Co-ordinates of A and B respectively  :

$A=\left(-a+\sqrt{a^{2}+b^{2}},-p \sqrt{p^{2}+q^{2}}\right)$

$B=\left(-a-\sqrt{a^{2}+b^{2}},-P-\sqrt{p^{2}+q^{2}}\right)$

Hence, equation of circle is
$\left(x+a-\sqrt{a^{2}+b^{2}}\right)\left(x+a+\sqrt{a^{2}+b^{2}}\right)+\left(y+p-\sqrt{p^{2}+q^{2}}\right)\left(y+p+\sqrt{p^{2}+q^{2}}\right)=0$

$\Rightarrow(x+a)^{2}-a^{2}-b^{2}+(y+p)^{2}-p^{2}-q^{2}=0$

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

Also, radius of circle is $\sqrt{a^{2}+b^{2}+p^{2}+q^{2}}$

QUESTION 9

ABCD is a square whose side is a taking AB and AD as axes, prove that the equation of the circle circumscribing the square is $x^{2}+y^{2}-a(x+y)=0$

Sol :

ABCD is a square with side a units.

Let AB and AD represent the x-axis and the y-axis, respectively.

Thus, the coordinates of B and D are $(a, 0)$ and $(0, a),$ respectively.

The end points of the diameter of the circle circumscribing the square are B and D

Thus, equation of the circle circumscribing the square is
$(x-a)(x-0)+(y-0)(y-a)=0$ or $x^{2}+y^{2}-a(x+y)=0$

QUESTION 10

The line $2 x-y+6=0$ meets the circle $x^{2}+y^{2}-2 y-9=0$ at A and B .Find the equation of the circle on AB as diameter.

Sol :

The equation of the line can be rewritten as $x=\frac{y-6}{2}$

Substituting the value of $x$ in the equation of the circle, we get:

$\Rightarrow \left(\frac{y-6}{2}\right)^{2}+y^{2}-2 y-9=0$

$\Rightarrow(y-6)^{2}+4 y^{2}-8 y-36=0$

$\Rightarrow y^{2}+36-12 y+4 y^{2}-8 y-36=0$

$\Rightarrow 5 y^{2}-20 y=0$

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

$\Rightarrow y(y-4)=0$

$\Rightarrow y=0,4$

At $y=0, x=-3$

At $y=4, x=-1$

Therefore, the coordinates of A and B are $(-1,4)$ and $(-3,0)$

$\therefore$ Equation of the circle with AB as its diameter:

$(x+1)(x+3)+(y-4)(y-0)=0$

$\Rightarrow x^{2}+4 x+y^{2}-4 y+3=0$

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