On the plane, a right angle is given. A circle with its center located outside the given angle is tangent to the bisector of the right angle, intersects one of its sides at points $A$ and $B$, and the extension of the other side at points $C$ and $D$. $AB = \sqrt{7}, CD = 1$. Find the radius of the circle.
Let $K$ be the vertex of the given right angle, $M$ the point of tangency with the bisector, $O$ the center of the circle, $R$ its radius, $P$ and $Q$ the projections of point $O$ onto $AB$ and $CD$, and $L$ the intersection of $OM$ and $AB$. Then $OQ^2 = R^2 - 1/4$, $PL^2 = OP^2 = R^2 - 7/4$. Note that $OQ + OP = KP + LP = KL + 2LP = \sqrt{2}(ML + OL) = \sqrt{2}R$, or $\sqrt{R^2 - \frac{1}{4}} + \sqrt{R^2 - \frac{7}{4}} = \sqrt{2}R$, $\sqrt{\left(R^2 - \frac{1}{4}\right)\left(R^2 - \frac{7}{4}\right)} = 2$, $\left(4R^2 - 1\right)\left(4R^2 - 7\right) = 16$, $4R^2 = 9$.

![](https://cdn.mathpix.com/cropped/2024_05_06_ae1b0d3bbbf9aa5c38e8g-48.jpg?height=346&width=334&top_left_y=1736&top_left_x=862)

## Answer

$1.5$.