10.307. A circle is drawn through two adjacent vertices of a square such that the length of the tangent to it, drawn from the third vertex, is three times the side of the square. Find the area of the circle if the side of the square is $a$.
## Solution.

Let $ABCD$ be the square mentioned in the problem (Fig. 10.96). The given circle passes through its vertices $A$ and $D$, $CE$ is a tangent, $AD=CD=a, CE=3a$. Extend $CD$ to intersect the circle at point $K$. Let $DK=x$. Then $CK \cdot CD = CE^2$; $(x+a) a = 9a^2$; $x = 8a$. Since $\angle ADK = 90^\circ$, $AK$ is the diameter of the circle and $AK^2 = AD^2 + DK^2$. Let $R$ be the radius of the given circle. Then $4R^2 = a^2 + 64a^2$; $R^2 = \frac{65a^2}{4}$. The area of the circle $S = \pi R^2 = \frac{65\pi a^2}{4}$.

Answer: $\frac{65\pi a^2}{4}$.