8,9

In an isosceles right triangle, the radius of the inscribed circle is 2.

Find the distance from the vertex of the acute angle to the point where the inscribed circle touches the opposite leg of this angle.
Let $a$ be the length of the leg. Then $2 \cdot 2 = 2a - a\sqrt{2}$, from which $a = 2\sqrt{2}(\sqrt{2} + 1)$. The square of the desired distance is $a^2 + 2^2 = 4(7 + 4\sqrt{2})$.

## Answer

$2 \sqrt{7 + 4\sqrt{2}}$.