Here's a (not-to-scale) picture of the situation:
Necessarily, each circle center ($D$, $E$, or $F$) is the point where an angle bisector meets an opposite edge; moreover, the points of tangency of a circle with the adjacent edges (for instance, $D^\prime$ and $D^{\prime\prime}$) are simply the feet of perpendiculars from the center to those edges.
We'll write $a$, $b$, $c$ for the lengths of sides $\overline{BC}$, $\overline{CA}$, $\overline{AB}$, and $d$, $e$, $f$ for the radii of $\bigcirc D$, $\bigcirc E$, $\bigcirc F$. Now, observing that each angle bisector cuts the triangle with into sub-triangles with convenient "bases" and "heights", we can compute the area, $T$, of $\triangle ABC$ in three ways:
$$T \;=\; \frac12 d\;(b+c) \;=\; \frac12 e\;(c+a) \;=\; \frac12 f\;(a+b) \tag{1}$$
Of course, writing $r$ for the inradius of $\triangle ABC$, we have a well-known fourth formula for area:
$$T \;=\; \frac12 r\;(a+b+c) \tag{2}$$
We can easily eliminate $a$, $b$, $c$ from the above. For instance,
$$\begin{align}
b+c = \frac{2T}{d}\quad c+a=\frac{2T}{e}\quad a+b = \frac{2T}{f} &\quad\to\quad 2(a+b+c) = 2T\left(\;\frac{1}{d}+\frac{1}{e}+\frac{1}{f}\;\right) \tag{3} \\
a+b+c = \frac{2T}{r} &\quad\to\quad 2(a+b+c) = 2T\left(\;\frac{2}{r}\;\right) \tag{4}
\end{align}$$
so that, as @Jack notes,
$$\frac{2}{r} = \frac{1}{d} + \frac{1}{e} + \frac{1}{f} \tag{$\star$}$$
Addendum (four years later!). As @jmerry has observed, the specific configuration in the problem statement is invalid. If we solve $(1)$ for $a$, $b$, $c$, we find
$$a = \left(-\frac1d + \frac1e + \frac1f \right)T \qquad b = \left(\frac1d - \frac1e + \frac1f \right)T \qquad c = \left(\frac1d + \frac1e - \frac1f \right)T$$
With $d=18$, $e=6$, $f=9$, these become $a=2T/9$, $b=0$, $c=T/9$, which do not correspond to a valid triangle ... not even a validly-degenerate one. (It's a good thing I didn't claim my picture was to scale.) A valid triangle requires that the three aspects of the Triangle Inequality hold
$$a \leq b+c \qquad b \leq c+a \qquad c \leq a+b$$
which, in turn, require
$$\frac3d \geq \frac1e + \frac1f \qquad \frac3e \geq \frac1f+\frac1d \qquad \frac3f \geq \frac1d+\frac1e$$
(The first of these is violated by the given values of $d$, $e$, $f$.)
Try to prove this..
•Find the length of $MA=2R\cos(\frac{A+2C}{2})$ first. ( Where $R$ is the circumradius of the triangle.)
•Then find $MB=2R\cos(\frac{A}{2})$ by using Sine Law (Chase the angles) in $\triangle MAB$
•Finally apply Pythagoras theorem in $\triangle MAQ$
$MQ^2-MA^2=MB^2-MA^2=AQ^2$ and $PQ=2AQ$
Best Answer
Let the feet of perpendiculars from point $A$, $B$ and $C$ on the respective opposite sides be $D$, $E$ and $F$ respectively. The circle with diameter $BH$ will pass through points $D$ and $F$. Let the tangents from $A$ and $C$ to this circle be $AS$ and $CT$ respectively.
Considering the power of points $A$ and $C$ with respect to this circle gives,
$AS^{2}=AF\cdot AB$
$CT^{2}=CD\cdot BC$
Now, observe that, $CD\cdot BC=CE\cdot AC$ because $\triangle ACD\sim \triangle BCE$. Similarly, $AF\cdot AB=AE\cdot AC$ because $\triangle CAF\sim \triangle BAE$.
$\Rightarrow AS^{2}=AF\cdot AB=AE\cdot AC$ and similarly $CT^{2}=CE\cdot AC$.
Adding these two equations gives, $AC^{2}=\boxed {AS^{2}+CT^{2}}$
Now plugging in the lengths of the tangents will give the value of $AC^{2}$.