6. Circle $k_{1}$ with a radius of 8 cm lies inside circle $k$. Both circles intersect circle $k_{2}$ with a radius of 15 cm, as shown in the figure. What is the radius of $k$ if the shaded area inside $k$ but outside $k_{1}$ is equal to the total shaded area inside $k_{2}$?

![](https://cdn.mathpix.com/cropped/2024_05_06_b124656272d750e4349eg-3.jpg?height=611&width=740&top_left_y=1091&top_left_x=681)
Answer: 17.

Solution. Let the area inside circle $k_{1}$ but outside circle $k_{2}$ be denoted by $T_{1}$; inside circle $k_{2}$ but outside circle $k$ by $T_{2}$; inside circle $k$ but outside circles $k_{1}$ and $k_{2}$ by $T_{3}$; and inside circles $k$ and $k_{2}$ but outside circle $k_{1}$ by $T$ (see the figure below). Additionally, let the radius of circle $k$ be denoted by $R$.

![](https://cdn.mathpix.com/cropped/2024_05_06_b124656272d750e4349eg-3.jpg?height=688&width=851&top_left_y=2043&top_left_x=611)

By the problem's condition, $T_{3}=T_{1}+T_{2}$. Let's write down the areas of circles $k_{2}$ and $k$: $T+T_{1}+T_{2}=15^{2} \pi, 8^{2} \pi+T+T_{3}=R^{2} \pi$. It remains to notice that $T+T_{3}=T+T_{1}+T_{2}=15^{2} \pi$, from which we get $R^{2} \pi=8^{2} \pi+15^{2} \pi, R^{2}=289, R=17$.

Remark. One could immediately notice that the difference between the sum of the areas of circles $k_{1}$ and $k_{2}$ and the area of circle $k$ equals the difference between the black and shaded areas.