Find the coordinates of the points of intersection of the circles

$$
(x-2)^{2}+(y-10)^{2}=50 \text { and } x^{2}+y^{2}+2(x-y)-18=0 .
$$
Let's find the coordinates of the intersection points $A\left(x_{1} ; y_{1}\right)$ and $B\left(x_{2} ; y_{2}\right)$ of the given circles by solving the system of equations

$$
\left\{\begin{array}{l}
(x-2)^{2}+(y-10)^{2}=50 \\
x^{2}+y^{2}+2(x-y)-18=0
\end{array}\right.
$$

We will get: $x_{1}=3, y_{1}=3, x_{2}=-3, y_{2}=5$.

## Answer

$(3 ; 3),(-3 ; 5)$.