How to find the equation for the circle when... A circle goes trough two points, $A=(-1,2)$ and $B=(3,0$). 
You also know that the center of the circle is an element of the following linear equation: $$k \leftrightarrow 2x+y+3=0 .$$
How can you find the equation of the circle when this is the only information? 
 A: HINT:
WLOG, the coordinate of the circle can be written as $O(k,-2k-3)$
Now $|OA|=|OB|$ both being the radius
A: EDIT: the center of the circle in general does NOT have to go through the line interpolating the two points $A$ and $B$. I assumed for some reason that $A$ and $B$ were on opposite sides of the circle, but there is no information in the problem actually indicating that.
Thus @lab bhattarchajee's answer below is the correct approach. I'll flesh out the details here for the sake of completeness.
Essentially we want to solve for the center of the circle, and we know that its coordinate must be of the form $$(-2z-3,z) $$ since it lies on the line $$y = -2x-3.$$
Now we are given two points on the circle, and by the definition of a circle we know that both points must have the same distance from the center, the radius. We will use this fact to create an equation which will allow us to solve for $z$, and hence the center of the circle.
$$dist(A,center)=dist(B,center)$$
Substituting the value of $z$ back into the formula we have found, we will also get the radius of the circle, and hence enough information to give the equation characterizing the circle.
In what follows I will square both sides of the equation to make the math simpler, i.e. I will solve: 
$$(dist(A,center))^2 = (dist(B,center))^2$$
In so doing I might introduce an additional solution to the equation, but this doesn't matter since we know that the radius of a circle must be positive, and therefore we can discard any extraneous negative solutions automatically. 
$$(dist(A,center))^2 = (-1 +2z+3)^2 +(2-z)^2$$
$$=(2z+2)^2 + 4-4z+z^2 = 4z^2 + 8z +4 +4 -4z +z^2 = 5z^2 +4z +8 $$
$$(dist(B,center))^2 = (3+2z+3)^2 + z^2 = 36+ 24z +5z^2$$
$$\implies 5z^2 +4z +8 = 36+ 24z +5z^2$$
$$\implies 20z = 8-36= -28 \implies z = -\frac{7}{5}$$
Therefore the center of the circle is
$$(-\frac{1}{5}, -\frac{7}{5})$$
and the square of the radius is
$$ \frac{61}{5}$$
Therefore the equation of the circle is
$$\frac{61}{5} = \left(x+\frac{1}{5}\right)^2 + \left(y+\frac{7}{5}\right)^2$$
Sorry again for the mistake from earlier.
