Intersection of two circles

By the Pythagorean Theorem, in triangle $\Delta ABC$

$\left( \frac 1 r \right) ^2 = r^2 + 1^2 \Leftrightarrow \frac 1 {r^2} = r^2 + 1 \Leftrightarrow 1 = r^4 + r^2 \Leftrightarrow r^4 + r^2 - 1 = 0$ .

This is actually a quadratic equation in terms of $r^2$ , so we can substitute in $x = r^2$ to get

$x^2 + x -1 = 0$ .

We can solve this by using the quadratic formula, so

$x = \frac {-b \pm \sqrt{b^2 - 4ac}} {2a} = \frac {-1 \pm \sqrt{1^2 - 4(1)(-1)}} {2(1)} = \frac {\sqrt{5} -1} {2}$ .

Since $x$ was defined to be $r^2$ , we can get $r$ by taking the square root of $x$ , but we can ignore the negative square root.

So $r = \sqrt{\frac {\sqrt{5} -1}{2}} = \frac 1 {\sqrt{\phi}}$ .

This means that $r$ is approximately $\boxed{0.786}$ .