Question

A circle is tangent to the x and y axes in the first quadrant at the points $P$ and $Q$ respectively. $BC$ and $AD$ are parallel tangents to the circle with slope $-1$. If the points $A$ and $B$ are on the y- axis while $C$ and $D$ are on the x-axis and the area of the quadrilateral $ABCD$ is $900\sqrt{2}$ sq. units, then the radius of the circle is

• A. $10$
• B. $12$
• C. $15$
• D. $20$

Answer 1

The correct option is C $15$


Radius $=\alpha$, $\alpha \in R$
Centre $=(\alpha ,\alpha )$
$MN\perp BC$, $MN\perp AD$
Equation of $MN$ is $y=x$
Equation of $MN$ in distance form is
$\frac{x-\alpha }{\cos 45}=\frac{y-\alpha }{\sin {45}^{\circ }}=\alpha$
$\Rightarrow x=\alpha +\frac{\alpha }{\sqrt{2}}=y$
$\therefore$ Coordinates of $M$ is $(\alpha +\frac{\alpha }{\sqrt{2}},\alpha +\frac{\alpha }{\sqrt{2}})$
Similarly, coordinates of $N$ is $(\alpha -\frac{\alpha }{\sqrt{2}},\alpha -\frac{\alpha }{\sqrt{2}})$
$x$ intercept of AD is $2\alpha +\sqrt{2}\alpha$
$x$ intercept of BC is $2\alpha -\sqrt{2}\alpha$
Area of trapezium ABCD $=\frac{1}{2}(AD+BC)MN$
$\Rightarrow 900\sqrt{2}=\frac{1}{2}\times \sqrt{2}\left(4\alpha \right)\times 2\alpha \Rightarrow \alpha =15$