A circle touches sides $AB$ and $AD$ of rectangle $ABCD$ at P and Q respectively and passes through vertex $C$. If distance of  C from chord $PQ$ is 5 units, then area of rectangle is:

Let AB=a and CB=b which are along axes
Let point C be (a,b)
Equation of PQ is x+y=h where (h,h) is centre of circle
Distance of C from  $\mathrm{PQ}=\frac{|\mathrm{a}+\mathrm{b}-\mathrm{h}|}{\sqrt{2}}=5$
$\Rightarrow {\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{h}}^2+2\mathrm{ab}-2\mathrm{ah}-2\mathrm{bh}=50...........\left(\mathrm{i}\right)$
Also distance of C from centre
$\mathrm{h}=\sqrt{{(\mathrm{a}-\mathrm{h})}^2+{(\mathrm{b}-\mathrm{h})}^2}$
${\mathrm{a}}^2+{\mathrm{b}}^2-2\mathrm{ah}-2\mathrm{bh}+{\mathrm{h}}^2=0.........\left(\mathrm{ii}\right)$
Solving eqns (i) and (ii) ab=25