Dear Student,
The square ABCD has circle C1 inscribed in it.
Let
O1E is the radius of C1. As C1 is inscribed in square ABCD, diameter of C1 = 1.
Therefore radius O1E = 1/2.
AC is the diagonal of square ABCD. AC = √2. Therefore O1C = √2/2.
Now we have e = EC = O1C – O1E = √2/2 – 1/2 = (√2 – 1)/2.
From the property of length of tangents, we have FC2 = GC x EC.
Let r be the radius of smaller circle C2. Then we have FC = r.
GC = EC – EG = e – 2r.
Hence we have r2 = (e – 2r)e. Which gives us r = (√2 – 1)e = (√2 – 1)2/2.
Area of C2 = πr2 = π(√2 – 1)4/4.
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)
