A circle with its center located inside a right angle touches one side of the angle, intersects the other side at points $A$ and $B$, and intersects the angle bisector at points $C$ and $D$. Given $AB = \sqrt{6}$, $CD = \sqrt{7}$. Find the radius of the circle.
Let $K$ be the vertex of the given right angle, $M$ the point of tangency, $O$ the center of the circle, $R$ its radius, $P$ and $Q$ the feet of the perpendiculars dropped from $O$ to $AB$ and $CD$, and $F$ the intersection of $OM$ and $CD$. Then $MK^2 = OP^2 = R^2 - \frac{3}{2}$, $OQ^2 = R^2 - \frac{7}{4}$, $MK = MF = MO - OF = MO - OQ \sqrt{2}$, or $\sqrt{R^2 - \frac{3}{2}} = R - \sqrt{2R^2 - \frac{7}{2}}$. From this, $R - \sqrt{R^2 - \frac{3}{2}} = \sqrt{2R^2 - \frac{7}{2}}$, $2R \sqrt{R^2 - \frac{3}{2}} = 2$, $R^2 = 2$.

![](https://cdn.mathpix.com/cropped/2024_05_06_36c4956010bd91c4b4eeg-16.jpg?height=297&width=492&top_left_y=229&top_left_x=789)

## Answer

$\sqrt{2}$.