Problem 11.5. Points $A, B, C, D, E, F$ are arranged clockwise on a circle, as shown in the figure. Chords $A D$ and $C E$ intersect at point $X$ at a right angle, and chords $A D$ and $B F$ intersect at point $Y$.

It is known that $C X=12, X E=27, X Y=15, B Y=9, Y F=11$.

![](https://cdn.mathpix.com/cropped/2024_05_06_20f31b6c2b9751326bf1g-4.jpg?height=506&width=495&top_left_y=716&top_left_x=479)

(a) (2 points) Find the length of segment $A D$.

(b) (2 points) Find the radius of the circle.
# Answer:

(a) (2 points) 36.

(b) (2 points) 19.5.

Solution. Let $A Y=a, X D=b$ (Fig. 8). We use the fact that the products of the segments of intersecting chords through a given point inside a circle are equal. This means that $324=12 \cdot 27=a(15+b)$ and $99=9 \cdot 11=b(a+15)$. Subtracting the second equation from the first, we get $225=15(a-b)$, which means $a-b=15$. Then $18^{2}=324=a(15+b)=(15+b)^{2}$, from which we obtain $18=15+b$ and $b=3$. Therefore, $a=18$ and $A D=a+15+b=3+15+18=36$. Thus, $A X=X D=18$. The center of the circle lies on the perpendicular bisector of the chord $A D$, that is, on the segment $C E$. Therefore, $C E$ is the diameter of the given circle, so its radius

![](https://cdn.mathpix.com/cropped/2024_05_06_20f31b6c2b9751326bf1g-5.jpg?height=502&width=497&top_left_y=92&top_left_x=478)

Fig. 8: to the solution of problem 11.5

is $\frac{1}{2} C E=\frac{1}{2}(12+27)=19.5$.