# Solving Tangential Circles Inside a Square

Facebook for me is just another platform to learn new things in Mathematics. If you are in the right group, you’re not just sharpen your math skills, it will make you more addicted to solving hard math problems.

Today, we will solve one of the questions I found very interesting to work on. This problem is about two circles tangent externally and inscribed in a square.

Problem: Courtesy of Vincent Earl Ryan Ty of EMC $ABCD$  is a square with side 10 units. Circle E is tangent to four sides of the square. Circle G is another circle tangent to two sides of the square and circle E. Find the area of circle G.

Solution:

Take note that points E, G, and D are collinear. Also observe that $\triangle FED$  is an isosceles right triangle with legs of 5 units. Using Pythagorean theorem in $FED$ we have, $DE^2=DF^2+EF^2$ $DE^2=5^2+5^2$ $DE^2=50$ $\sqrt{DE^2}=\sqrt{50}$ $DE=5\sqrt{2}$

Also observe that $\triangle DGM$ is an isosceles right triangle. Thus $DG=MG\sqrt{2}$. We know that $MG=r$ or equal to the radius of the small circle. The one we need to find its area. Now, we just have to find $DG$.

Solving for DG: $DG=DE-EG$ $DG=5\sqrt{2}-(5+r)$ $DG=5\sqrt{2}-5-r$

Now, going back to $DG=MG\sqrt{2}$  we have, $DG=MG\sqrt{2}$ $5\sqrt{2}-5-r=r\sqrt{2}$

Regrouping the terms we have, $5\sqrt{2}-5=r\sqrt{2}+r$

Factor out common terms, $5(\sqrt{2}-1)=r(\sqrt{2}+1)$ $r=\displaystyle\frac{5(\sqrt{2}-1)}{\sqrt{2}+1}$

Rationalizing the denominator we have, $r=\displaystyle\frac{5(\sqrt{2}-1)}{\sqrt{2}+1}\cdot \displaystyle\frac{\sqrt{2}-1}{\sqrt{2}-1}$ $r=\displaystyle\frac{5(\sqrt{2}-1)^2}{(\sqrt{2})^2-1^2}$ $r=\displaystyle\frac{5(\sqrt{2}-1)^2}{2-1}$ $r=5(\sqrt{2}-1)^2$ $r=5(2-2\sqrt{2}+1)$ $r=5(3-2\sqrt{2})$ units

Solving for the Area of the circle we have, $A_\circ=\pi r^2$ $A_\circ=\pi (5(3-2\sqrt{2}))^2$ $A_\circ=\pi 25(9-12\sqrt{2}+8)$ $A_\circ=25(17-12\sqrt{2})\pi$ sq. units