I'm interpreting your question as follows: Find the locus of all $P$ such that if tangents are drawn from $P$ to circle $A$, then the angle between those tangents will be the same as the angle between the tangents drawn from point $P$ to circle $B$.
If I am interpreting your question correctly, you're correct in assuming that the locus is a circle. I'm going to use a geometric proof to show you why.
Call the center of one circle $A$ and the other $B$. Call the radius of the first circle $r_a$, and the radius of the second $r_b$. W.L.O.G. $r_a<r_b$. Mark the intersection of the internal tangents of the circles $I$. Mark the intersection of their external tangents $E$. Call the midpoint of $I$ and $E$ point $K$. Call the circle centered at $K$ with diameter $IE$ circle $O$. Then, the locus of points in your problem is circle $O$.
Proof:
Claim: If $P$ is in the locus, then it is on circle $O$.
Justification: $P$ is on the locus if and only if $\displaystyle \frac{AP}{BP}=\frac{r_a}{r_b}$(can you figure out why?)
Using law of sines, you can also figure out that $\displaystyle \frac{AP}{BP}=\frac{AI}{BI}\cdot\frac{\sin\left(\angle IPB\right)}{\sin\left(\angle IPA\right)}$
Since $\displaystyle \frac{AI}{BI}=\frac{r_a}{b_a}$, we have that $\displaystyle \frac{\sin\left(\angle IPB\right)}{\sin\left(\angle IPA\right)}=1$
Because $\angle IPA + \angle IPB <\pi$, we have $\angle IPA = \angle IPB$
Similarly, using law of sines, you can prove that $\displaystyle \frac{AP}{BP}=\frac{AE}{BE}\cdot\frac{\sin\left(\angle EPB\right)}{\sin\left(\angle EPA\right)}$.
Since $\displaystyle \frac{AE}{BE}=\frac{r_a}{r_b}$, we have that $\displaystyle \frac{\sin\left(\angle EPB\right)}{\sin\left(\angle EPA\right)}=1$
Because $r_a<r_b$, we have that $\angle EPA<\angle EPB$. Ergo, $\angle EPA=\pi-\angle EPB$.
Using what we know, let's try to calculate $\angle IPE$:
\begin{align}
\angle IPE &= \angle IPA+\angle EPA\\
&=\angle IPA+\pi-\angle EPB\\
&=\angle IPA+\pi-\left(\angle EPA+\angle BPA\right)\\
&=\angle IPA+\pi-\left(\angle EPA+2\angle IPA\right)\\
&=\pi-\left(\angle EPA + \angle IPA\right)\\
&=\pi-\angle IPE
\end{align}
Ergo $\angle IPE=\displaystyle \frac{\pi}{2}$. This implies that $P$ is on the circle with diameter $IE$. In other words, $P$ is on circle $O$.
Claim: If $P$ is on circle $O$, then $P$ is in the locus.
Justification:
Using law of sines, you can derive the following equations:
$$\frac{\sin\left(\angle IPA\right)}{\sin\left(\angle IPB\right)}=\frac{AI}{BI}\cdot\frac{BP}{AP}=\frac{r_a}{r_b}\cdot\frac{BP}{AP}$$
$$\frac{\sin\left(\angle EPA\right)}{\sin\left(\angle EPB\right)}=\frac{AE}{BE}\cdot\frac{BP}{AP}=\frac{r_a}{r_b}\cdot\frac{BP}{AP}$$
Ergo, we have the following:
$$\frac{\sin\left(\angle IPA\right)}{\sin\left(\angle IPB\right)}=\frac{\sin\left(\angle EPA\right)}{\sin\left(\angle EPB\right)}$$
We can rearrange this equation like this:
$$\frac{\sin\left(\angle IPA\right)}{\sin\left(\angle EPA\right)}=\frac{\sin\left(\angle IPB\right)}{\sin\left(\angle EPB\right)}$$
We also know that $\displaystyle \angle IPE = \angle EPA + \angle IPA = \angle EPB-\angle IPB =\frac{\pi}{2}$.
So, we can make substitutions to get the following:
$$\frac{\sin\left(\angle IPA\right)}{\sin\left( \frac{\pi}{2}-\angle IPA\right)}=\frac{\sin\left(\angle IPB\right)}{\sin\left(\frac{\pi}{2}+\angle IPB\right)}$$
We can simplify this expression like so:
$$\frac{\sin\left(\angle IPA\right)}{\cos\left( \angle IPA\right)}=\frac{\sin\left(\angle IPB\right)}{\cos\left(\angle IPB\right)}$$
We can again simplify to get the following:
$$\tan\left(\angle IPA\right)=\tan\left(\angle IPB\right)$$
Because $\angle IPA + \angle IPB < \pi$, we have $\angle IPA = \angle IPB$.
Again, we use the the previous equation:
$$\frac{AP}{BP}=\frac{r_a}{r_b}\cdot\frac{\sin\left(\angle IPB\right)}{\sin\left(\angle IPA\right)}$$
We can simplify this using $\angle IPA = \angle IPB$:
$$\frac{AP}{BP}=\frac{r_a}{r_b}\cdot 1=\frac{r_a}{r_b}$$
This implies that $P$ is in the locus.
Since we proved that the circle is a subset of the locus, and the locus is a subset of the circle, we have that the locus and the circle are one and the same.
Best Answer
Let $C(3,4),D(1,2)$. Also, let $E(X,Y)$ be the midpoint of $AB$.
$\qquad\qquad\qquad$
Since $\triangle{CAE}$ is a right triangle with $$|AC|=6,\quad |CE|=\sqrt{(X-3)^2+(Y-4)^2}$$ we have $$|AE|^2=|AC|^2-|CE|^2=36-(X-3)^2-(Y-4)^2\tag1$$
Also, since we can see that $D$ is on the circle whose diameter is the line segment $AB$, we have $$|DE|=|AE|\tag 2$$
From $(1)(2)$, we have $$36-(X-3)^2-(Y-4)^2=(X-1)^2+(Y-2)^2,$$ i.e. $$X^2+Y^2-2\cdot 2X-2\cdot 3Y-3=0.$$
Hence, $a=2,b=3,c=3.$