If the radius of the smallest circle is x cm form an equation in x and show that it reduces to a quadratic equation

Question

What i tried

I tried finding the area of the rectangle ABCD which is $5*7=35$ and then i tried to find the area of the three quadrants in terms of $x$ and add those areas up, however, i still couldn't find the area of the two region that arent under the quadrants. I have tried various ways to find the area of these two regions outside the quadrants but i couldn't find them. Could anyone explain please. Thanks

Answer 1

Hint:

1. The radius of the upper left circle is $5\,\mathrm{cm}$ and that of the lower right circle is $(5-x)\,\mathrm{cm}$

2. Consider the diagonal $AC$ of the rectangle.

Answer 2

Let $x$ be the radius of top right circle. Then $5 - x$ is radius of bottom right circle.

$((5 - x) + 5)^2 = 7^2 + 5^2 = \text{square of length of diagonal $AC$}$

$100 + x^2 - 20x = 49 + 25$

$x^2 -20x +26 = 0$

Joining centers of circles to the common point of intersection of top left and bottom right circle is a straight line because tangents are perpendicular to the line joining center to point of contact of the tangent