I am given that $P(-3,-1)$ and $Q(5,3)$ are points of the circle. Also, the line $L:0=x+2y-13$ is tangent to said circle. The objective is to find the equation of the circle.
I thought of a way for solving this, but it doesn't seem to be the best one. The option was to find the points equidistant to $P$ and $Q$. Afterwards, using any point $(x,y)$ of the line of points equidistant to $P$ and $Q$, the next step would be to make the distance from $(x,y)$ to $L$ the same as to the distance to $P$ or $Q$.
Equations: $$ 2a + b = 3 \\ 5a^2-10a+25 = r^2\\ \sqrt{5}r = a+2b-13 $$
HINT:
If $(a,b)$ is the center of the required circle
we have $$(a+3)^2+(b+1)^2=(a-5)^2+(b-3)^2=r^2$$ where $r$ is the radius
Now, the perpendicular distance of the tangent form the center $(a,b)$ is again equal to the radius
Do you know how to calculate the perpendicular distance?