The equation of a circle is given as:

$$
x^{2}+y^{2}=25
$$

Let's draw tangents from the point $P(2,-8)$ to the circle, and let $A$ and $B$ be the points of tangency. Calculate the volume of the solid formed by rotating the triangle $A B P$ around the $X$-axis.
The coordinates of the point of tangency satisfy the following relationship:

$$
\begin{gathered}
x^{2}+y^{2}=25 \\
x_{0} x+y_{0} y=25
\end{gathered}
$$

where $x_{0}, y_{0}$ are the coordinates of point $P$.

From the two equations, the points $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$ can be determined:

$$
\begin{array}{cc}
x_{1}=4.58 & x_{2}=-3.14 \\
y_{1}=-1.98 & y_{2}=-3.91
\end{array}
$$

Since the volume of the generated solid of revolution is

$$
K=2 \pi d T
$$

where $d$ is the distance of the centroid from the axis of rotation (in this case, its ordinate), and $T$ is the area of the triangle. The numerical value of the volume is obtained by calculating the following expression and ignoring the sign:

$$
K=\pi \frac{y_{0}+y_{1}+y_{2}}{3}\left|\begin{array}{lll}
x_{0} & y_{0} & 1 \\
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1
\end{array}\right|
$$

The calculation of this product gives the volume:

$$
K=603.5
$$

The problem was solved by: Lukhaub Gy., Prohászka, Weisz A.