Question

A circle touches the sides $AB$ and $AD$ of a rectangle $ABCD$ at $P$ and $Q$ respectively and passes through the vertex $C$. If the distance of $C$ from $PQ$ is $97$ units, then area of the rectangle in sq. units is

Answer 1

Let $ABCD$ be a rectangle with $A(a,0)$,$B(a,b)$,$C(a,0)$,$D(0,0)$.
Given $AP$ and $AQ$ are tangents to circle.
$\Rightarrow AP=AQ$ (Tangents drawn from an external point are equal)
Let $AQ=m$
$\Rightarrow QD=b-m$
So, coordinates of $Q$ are $(0,b-m)$.
Coordinates of $P$ are $(m,b)$ ($\because AP=m$)
Now, since,$y-axis(x=0)$ is tangent to circle.
So, the center of circle is $(m,b-m)$
So, equation of circle is
$(x-m{)}^2+(y-b+m{)}^2=m^2$
$\Rightarrow x^2+(y-b+m{)}^2-2mx=0$
Since, the circle passes through $C(a,0)$
$\Rightarrow a^2+(m-b{)}^2-2ma=0$ .....(1)
Equation of $PQ$ is
$x-y=m-b$
Now, perpendicular distance of $C(a,0)$ from $PQ$ is
$|\frac{a-m+b}{\sqrt{2}}|=97$
Squaring both sides , we get
$(a-(m-b){)}^2=2(9409)$
$a^2+(m-b{)}^2-2am+2ab=2(9409)$
$\Rightarrow ab=9409$ (by (1))
Hence, area of rectangle is $9409$ sq.units