$\left\{\begin{array}{l}x^{2}+y^{2}=25 \\ x+2 y=5\end{array}\right.$

Step 1 ：We are given a system of two equations: \(x^{2}+y^{2}=25\) and \(x+2y=5\). The first one is a circle equation and the second one is a linear equation.

Step 2 ：We can solve this system of equations by substitution or elimination method. Here, we will use the substitution method.

Step 3 ：First, we solve the second equation for x, which gives us \(x = 5 - 2y\).

Step 4 ：We then substitute \(x = 5 - 2y\) in the first equation. This gives us a quadratic equation in terms of y: \(y^{2} + (5 - 2y)^{2} = 25\).

Step 5 ：Solving this quadratic equation gives us the values of y: \(y = 0\) and \(y = 4\).

Step 6 ：We substitute these values of y in the second equation to get the corresponding values of x: \(x = 5\) when \(y = 0\) and \(x = -3\) when \(y = 4\).

Step 7 ：The solutions to the system of equations are \(\boxed{(x, y) = (5, 0)}\) and \(\boxed{(x, y) = (-3, 4)}\).