10.195. The circle touches two adjacent sides of the square and divides each of the other two sides into segments equal to 2 and 23 cm. Find the radius of the circle.
## Solution.

Let $M$ and $K$ be the points of tangency of the circle with the sides $AB$ and $AD$ of the square $ABCD$ (Fig. 10.5), $M$ and $K$ being the points of intersection of the circle with the specified sides of the square, and $P$ and $E$ being the points of intersection with the sides $BC$ and $CD$. By the problem statement

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0713.jpg?height=483&width=502&top_left_y=126&top_left_x=177)

Fig. 10.5

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0713.jpg?height=551&width=396&top_left_y=76&top_left_x=785)

Fig. 10.6

$BP = DE = 2$ cm, $CP = CE = 23$ cm. Draw $ON \perp BC$. $OM = OP = NB = OK = R$, where $R$ is the radius of the given circle. $AB = NK = BC = CP + PB = 25$ cm, $ON = NK - KO = (25 - R)$ cm, $PN = BN - BP = (R - 2)$ cm. From the right triangle $PNO$: $OP^2 = ON^2 + NP^2$ or $R^2 = (R - 2)^2 + (25 - R)^2$, from which $\left[\begin{array}{l}R=17 \\ R=37\end{array}\right.$. The condition is satisfied by $R = 17$.

Answer: 17 cm.