Question

A circle touched the sides AB and AD of a rectangle at P and Q respectively and passes through the vertex C . If the distance of C from PQ is 5 units . then prove that the area of the rectangle is 25 sq. unit

Answer 1

Let the sides AB and AD be chosen along the axes and hence
the circle touches the axes so that it is (h,h),h
$(x-h{)}^2+(y-h{)}^2=h^2$
Also equation of PQ by intercepts form is
$\frac{x}{h}+\frac{y}{h}$ = 1 or x + y = h
as AP=AQ=h
Let the vertex C be (a, b) whose distance from PQ is 5,
given
$\therefore \frac{a+b-h}{\sqrt{2}}=5$
$\therefore a+b=5\sqrt{2}+h$
We have to find the area of rectangle, i.e., ab
Again (a,b) lies on circle (1)
$\therefore (a-h{)}^2+(b-h{)}^2=h^2$
$a^2+b^2-2h(a+b)+h^2=0$
Putting the value of h from (3),
$(a^2+b^2)-2(a+b)[a+b-5\sqrt{2}]+(a+b-5\sqrt{2}{)}^2=0$
or $(a+b{)}^2-2ab-2(a+b{)}^2+10\sqrt{2}(a+b)+(a+b{)}^2-10\sqrt{2}(a+b)+50=0$
or -2ab + 50 = 0
$\therefore$ ab = 25 = area of rectangle