THIS ANSWER GIVES YOU THE FAMILY OF CIRCLES THROUGH TWO GIVEN POINTS.
The way I would work is something like this... if the two points $P_1(x_1,y_1)$ and $P_2(x_2,y_2)$ are on your circle then the line segment $[P_1P_2]$ is a chord. Suppose that the distance $|P_1P_2|=2d$.
Now the perpendicular from the centre of a circle to a chord bisects the chord. We can get the equation of this perpendicular bisector because we have that the midpoint of $[P_1P_2]$ is on it... $((x_1+x_2)/2,(y_1+y_2)/2)$. The slope of the perpendicular bisector, $m$, satisfies $$m\cdot m_{[P_1P_2]}=-1,$$
so is
$$m=-\frac{x_2-x_1}{y_2-y_1},$$
so the perpendicular bisector has equation
$$y=-\frac{x_2-x_1}{y_2-y_1}x+\frac{x_2-x_1}{y_2-y_1}\frac{x_1+x_2}{2}+\frac{y_1+y_2}{2},$$
briefly $k=mh+c$ where $k\sim y$, $h\sim x$ and
$$m=-\frac{x_2-x_1}{y_2-y_1},$$
and
$$c=\frac{x_2-x_1}{y_2-y_1}\frac{x_1+x_2}{2}+\frac{y_1+y_2}{2}$$
Here $h$ is free --- each $h$ gives you a different centre $(h,k)=(h,mh+c)$.
By drawing a picture you will see a RAT triangle with vertices at, say the midpoint of $[P_1P_2]$, the centre $(h,mh+c)$ and $P_1$. Using Pythagoras we have that
$$r^2=d^2+\left(\frac{x_1+x_2}{2}-h\right)^2+\left(\frac{y_1+y_2}{2}-(mh+c)\right)^2$$
So we have that the circles in question are:
$$\{(x-h)^2+(x-(mh+c))^2=r^2:h\in\mathbb{R}\},$$
where each of $m$ and $c$ are functions of $P_1,\,P_2$ and --- once $h$ is chosen --- $r$ is a function of $P_1$, $P_2$ and $h$.
Best Answer
Take the centre $O$ of a circle through two points $AB$. Then $OA=OB$ because they are radii of the same circle. Hence $OAB$ is isosceles, and the altitude from $O$ to point $P$ on $AB$ defines two congruent triangles $OPA$ and $OPB$. So $OP$ is the perpendicular bisector of $AB$.
Any point $O$ on the perpendicular bisector likewise defines two congruent triangles so that $OA=OB$.