Example 22 A circle is tangent to the line $2 x-y+5=0$ at point $M(-2,1)$, and passes through point $N(3,2)$. Find the equation of this circle.
Solve: Represent point $M$ in the form of a point circle: $(x+2)^{2}+(y-1)^{2}=0$. Let the equation of the required circle be $(x+2)^{2}+(y-1)^{2}+\lambda(2 x-y+5)=0$.
Substitute the coordinates of point $N$ into the above equation, and solve to get $\lambda=-\frac{26}{9}$.
Thus, the equation of the required circle is $9 x^{2}+9 y^{2}-16 x+8 y-85=0$.