Problem of Apollonius: Find the radius of the circle tangent to $3$ other circles $O_1$, $O_2$ and $O_3$ have radius of $a$, $b$ and $c$ Find the radius of the circle tangent to $3$ other circles $O_1$, $O_2$ and $O_3$ have radius of $a$, $b$ and $c$
The Wikipedia page about the Problem of Apollonius can be found here
I can't provide you an exact image because the construction is too complex. What have I done is draws tons of lines to attempt to use Pythagoras's theorem and trig to solve it.

If you can endure the algebraic slog (a computer algebra system really helps), you can find some structure that simplifies the solution.

Let the given circles have radii $r_A$, $r_B$, $r_C$, and (non-collinear) centers $A:=(x_A,y_A)$, $B:=(x_B,y_B)$, $C:=(x_C,y_C)$; I specifically used
$$A = (0,0) \qquad B = (c,0) \qquad C = (b\cos A, b \sin A)$$
As Wikipedia's Problem of Apollonius entry notes, there are eight solutions that depend on whether each circle's tangency with the target circle is external or internal. The entry's coordinate discussion accommodates this by attaching sign variables to the radii; here, we can just assert that the radii $r_A$, $r_B$, $r_C$ are signed values. 
Let the target circle have center $P:=(x_P,y_P)$ and radius $r$. We can express $P$ in barycentric coordinates:
$$P = \frac{\alpha A+\beta B+\gamma C}{\alpha+\beta+\gamma} \tag{1}$$
Now, we'll follow Wikipedia's lead and consider the equations
$$\begin{align}
(x_P-x_A)^2 + (y_P-y_A)^2=(r-r_A)^2 \\
(x_P-x_B)^2 + (y_P-y_B)^2=(r-r_B)^2 \\
(x_P-x_C)^2 + (y_P-y_C)^2=(r-r_C)^2
\end{align} \tag{2}$$
Upon subtracting the latter two equations from the first, we eliminate troublesome quadratic terms, leaving a linear homogeneous system of two equations in the three unknowns $\alpha$, $\beta$, $\gamma$. The reader can verify that we can write the solution as 
$$\begin{align}
\alpha &= a^2 b c \cos A - a^2 r_A (r_A - 2 r) + a b \cos C\,r_B (r_B - 2 r) + c a \cos B\,r_C (r_C - 2 r) \\[4pt]
\beta  &= a b^2 c \cos B - b^2 r_B (r_B - 2 r) + b c \cos A\,r_C (r_C - 2 r) + a b \cos C\,r_A (r_A - 2 r) \\[4pt]
\gamma &= a b c^2 \cos C - c^2 r_C (r_C - 2 r) + c a \cos B\,r_A (r_A - 2 r) + b c \cos A\,r_B (r_B - 2 r)
\end{align} \tag{3}$$
where $a$, $b$, $c$ are the sides of $\triangle ABC$. Conveniently, $\alpha+\beta+\gamma = 16 |\triangle ABC|^2$, by Heron's Formula.
Substituting $P$ with these $\alpha$, $\beta$, $\gamma$ values back into the first equation of $(1)$, we get this quadratic in $r$:
$$\begin{align}
0 &= 4 r^2 \left(\;-4|\triangle ABC|^2 
 + \sum_{cyc} \left(a^2 r_A^2 - 2 b c \cos A\,r_B r_C \right) \;\right) \\
&-4r \left(\;
  \sum_{cyc}\left( a^2 r_A^3 - b c \cos A (a^2 r_A + r_B^2 r_C + r_B r_C^2) \right)
  \;\right) \\[4pt]
&+a^2 b^2 c^2 + \sum_{cyc} \left( a^2 r_A^4 - 2 b c \cos A (a^2 r_A^2 + r_B^2 r_C^2)\right)
\end{align} \tag{8}$$
Somewhat surprisingly, the discriminant simplifies nicely:
$$\triangle = 64 |\triangle ABC|^2 \left(a^2 - (r_B - r_C)^2\right)\left( b^2 - (r_C - r_A)^2\right) \left(c^2 - (r_A - r_B)^2\right) \tag{9}$$
The last three factors invite defining, say, $d$, $e$, $f$ such that
$$d^2 = a^2 - (r_B - r_C)^2 \qquad e^2 = b^2 - (r_C - r_A)^2 \qquad f^2 = c^2 - (r_A - r_B)^2 \tag{10}$$
Since the differences could be negative, we allow that $d$, $e$, $f$ could be imaginary; but $d^2e^2f^2$ must be non-negative (so, $def$ is real). As it turns out, it's helpful to define (possibly-imaginary) "angles" $D$, $E$, $F$ with
$$\cos D = \frac{-d^2+e^2+f^2}{2e f} \qquad \cos E = \frac{-e^2+f^2+d^2}{2fd} \qquad \cos F = \frac{-f^2+d^2+e^2}{2de} \tag{11}$$
and (possibly-imaginary) "area" $|\triangle DEF|$ with
$$|\triangle DEF|^2 = \frac1{16}(d+e+f)(-d+e+f)(d-e+f)(d+e-f) \tag{12}$$
How helpful? Well, helpful enough that $(8)$ shrinks to
$$\begin{align}
0 &= 16 r^2 |\triangle DEF|^2 - 4r d e f ( d r_A\cos D + er_B \cos E + fr_C \cos F) \\[4pt]
&-\left(d^2 e^2 f^2 + d^4 r_A^2 + e^4 r_B^2 + f^4 r_C^2 - 2 e^2 f^2 r_B r_C - 2 f^2 d^2 r_C r_A - 2 d^2 e^2 r_A r_B  \right)
\end{align} \tag{8'}$$
(If we defined possibly imaginary $s_A$, $s_B$, $s_C$ with $s_A^2 = d^2 r_A$, etc, the constant term could be written $(s_A+s_B+s_C)(-s_A+s_B+s_C)(s_A-s_B+s_C)(s_A+s_B-s_C)-d^2 e^2 f^2$, but that isn't particularly helpful going forward.)
Solving $(8')$ yields

$$r = \frac{def}{8|\triangle DEF|^2} \left(\;
    dr_A \cos D
     + e r_B \cos E 
     + f r_C \cos F
     \pm 2|\triangle ABC|
   \;\right) \tag{13}$$

Note that, if we were to multiply-through the $def$ factor, and expand the cosines via $(11)$, the first three terms would only involve even powers $d$, $e$, $f$, making for a real (possibly-negative) value. We know that $def$ is real, so the $\pm 2def\,|\triangle ABC|$ term is also real. This guarantees that $r$ itself is real.
Since the signs of $r_A$, $r_B$, $r_C$ account for the eight possible tangency configurations, presumably one choice of "$\pm$" is always extraneous. Can we decide which one ahead of time? Maybe, but I'm running out of steam ...
Substituting $(13)$ into a $def$ version of $(7)$, and dividing-through by a common factor, yields a final-ish form for parameter $\alpha$:

$$\begin{align}
\alpha = d 
\left(
\begin{array}{l}
\phantom{+} \cos D
\left(
4 |\triangle DEF|^2 + \sum_{cyc} \left( d^2 r_A^2 - 2 e f r_B r_C \cos D \right)
\right) \\
\pm 2 |\triangle ABC|\; (d r_A - e r_B \cos F - f r_C \cos E)
\end{array}
\right)
\end{align} \tag{14}$$

If we define $u := d r_A$, $v := e r_B$, $w := f r_C$, we reduce a bit of clutter in $(13)$ and $(14)$:
$$\begin{align}
r &= \frac{def}{8|\triangle DEF|^2} 
\left(\;u \cos D + v \cos E + w \cos F\pm 2|\triangle ABC|\;\right) \\
\alpha &= d 
\left(
\begin{array}{l}
\phantom{+} \cos D
\left(
4 |\triangle DEF|^2 + u^2 + v^2 + w^2 - 2 v w \cos D - 2 w u \cos E - 2 u v \cos F
\right) \\[6pt]
\pm 2 |\triangle ABC|\; (u - v \cos F - w \cos E)
\end{array}
\right)
\end{align} \tag{15}$$

As a final note, I'll mention that the values $d$, $e$, $f$ have geometric significance related to the power of certain points with respect to the given circles. It's a little tricky to describe, and I don't (yet) see how it might contribute to the discussion, so I won't bother including it.
