HINT:
Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$
If the circle passes through $A(a,d),B(b,e),C(p,q)$
$a^2+d^2+2g\cdot a+2f\cdot d+c=0$
$b^2+e^2+2g\cdot b+2f\cdot e+c=0$
$p^2+q^2+2g\cdot p+2f\cdot q+c=0$
So, we have three equations with three unknowns $g,f,c$
What can we conclude from here?
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
Given a circle with radius $CE$ and two points $A$, $B$ outside the circle.
Join $AB$ and set up the perpendicular bisector at $D$.
On this line lie the centers of all circles passing through $A$, $B$, and from each of these points lines can be drawn: to $A$, and through center $C$ to the opposite side of the given circle (or to center $C$ through the near side).
Let $F$ be the (unique) point on the perpendicular bisector such that$$FA=FCE$$The circle drawn with center $F$ and radius $FA$ is internally tangent to the given circle, since common point $E$ lies on the line through their centers but not between them.
Next let $G$ be the (unique) point equidistant from $A$ and point $H$ on the near side of the given circle. The circle with center $G$ and radius $GA$ is externally tangent to the given circle since now $H$ lies on the line through their centers and between them.
I have not shown how to construct points $F$ and $G$, but if it's clear that these points on the perpendicular bisector exist, then two and only two circles pass through points $A$, $B$ tangent to the given circle, one internally and one externally.