Question

Find the distance between center of a unit circle and an internally tangent circle,which tangent's meet at the outer's center in a 72 degree angle.

Answer 1

Let $m=\tan \left({36}^0\right)=\sqrt{5-2\sqrt{5}}$

We may supper that we have

$x^2+y^2=1$

$mx-y=0$

$mx+y=0$

We want to find a circle ${(x-a)}^2+y^2=r^2$ where $0<a<1$, $0<r<1$ which touches each of $\left(1\right)\left(2\right)\left(3\right)$

First $r=1-a$

Second since distance between $(a,0)$ and $\left(2\right)$ is $r$

We have

$r=\frac{|am-0|}{\sqrt{m^2+1^2}}$

From these, we get

$a=\frac{4}{4+\sqrt{10-2\sqrt{5}}}r=1-\frac{4}{4+\sqrt{10-2\sqrt{5}}}$

$a-r=\frac{8}{4+\sqrt{10-2\sqrt{5}}}-1=0.259616$.