A circle passing through the vertex C of a rectangle ABCD and touches its sides AB and AD at M and N respectively. If ${\perp }^{lar}$ distance from C to the line segment $\overline{MN}$ is equal to 5 then area of rectangle ABCD is K then the unit place of K is

Let r be the radius of the circle and A = (0, 0) AB is along x-axis and y-axis.
Equation of circle is ${(x-r)}^2+{(y-r)}^2=r^2$ --------------(1)

Equation of MN is  $x+y=r$
Now ${\perp }^{lar}$ distance from C(p, q) to the above line is 5
$\Rightarrow \left|\frac{p+q-r}{\sqrt{2}}\right|=5\Rightarrow {(p+q-r)}^2=50$ -------------(2) 
(p, q) lies on circle (1) $\Rightarrow p^2+q^2-2r(p+q)+r^2=0$
$$\begin{gathered} \Rightarrow {(p+q-r)}^2-2pq=0 \\ 50-2pq=0\Rightarrow pq=25 \end{gathered}$$ 
$\therefore$ Area of rectangle ABCD is 25