Length of tangent common to two semicircles which are also tangent to a larger semicircle. In the diagram,the semicircles centered at ${P}$ and $Q$ are tangent to each other and to the large semicircle ,and their radii are $6$ and $4$ respectively.Line $LM $ is tangent to semicircles $P$ and  $Q$ .Find $LM$

Efforts made:
I've been able to calculate the length of the segment between the points of tangency of the interior semicircles ,but i've not been able to find  the lengths of the external segments.
This problem is meant to be solved by synthetic methods(geometric methods).
thanks in advance
 A: DISCALIMER : This is a brute force method using coordinate geometry. A better method should exists.

Take the center of the large circle be $(0, 0)$. Then $P = (-4, 0)$ and $Q = (6, 0)$. 
Now let $y = mx + c$ denote the line extending $LM$ on both sides, where $m, c \in \mathbb{R}$ are constants such that both the systems of equations,
$$y = mx + c$$
$$(x - (-4))^2 + y^2 = 6^2$$
and
$$y = mx + c$$
$$(x - 6)^2 + y^2 = 4^2$$
have only one unique solution. 

From the first system we can work out, by substitution first and then by considering discriminant second, one required condition:
$$(2mc + 8)^2 - 4(m^2 + 1)(c^2 - 20) = 0$$
$$20 m^2 + 8mc - c^2 + 36 = 0$$
and similarly from the second system,
$$(2mc - 12)^2 - 4(m^2 + 1)(c^2 + 20) = 0$$
$$20m^2 + 12mc + c^2 - 16 = 0$$

Subtract the twice the second condition from thrice the first to eliminate $mc$:
$$3(20 m^2 + 8mc - c^2 + 36) - 2(20m^2 + 12mc + c^2 - 16)= 0$$
$$20m^2 -5c^2 + 140 = 0$$
$$c^2 = 4m^2 + 28$$

Now consider the sum of $4$ times the first condition and $9$ times the second condition to eliminate the constant term:
$$4(20 m^2 + 8mc - c^2 + 36) + 9(20m^2 + 12mc + c^2 - 16) = 0$$
$$52m^2 + 28mc + c^2 = 0$$
$$(2m + c)(26m + c)$$
From which we get either $c = -2m$ or $c = -26m$.

Using the results in the previous two parts, we see that there a few possibilities for $m$. Clearly $c = -2m$ has no solutions because
$$(-2m)^2 = 4m^2 + 28$$
has no solution for $m$. Hence $c = -26m$. Therefore
$$(-26m)^2 = 4m^2 + 28$$
$$672m^2 = 28$$
$$m = \pm \frac{1}{2\sqrt{6}}$$
Since from our geometric definition of $m$ being the gradient of the negatively sloped $LM$-extended, $m$ must be negative. Hence 
$$m = -\frac{1}{2\sqrt{6}}$$
which leaves $$c = \frac{13}{\sqrt{6}}$$
Thus, the line $LM$ extended has equation $y = -\frac{1}{2\sqrt{6}}x + \frac{13}{\sqrt{6}}$

Now use the equation of $LM$ extended and the equation of the large circle (with radius $\frac{6 + 6 + 4 + 4}{2} = 10$,
$$x^2 + y^2 = 10^2$$
to find the coordinates of $L$ and $M$.
Finding the distance between $L$ and $M$ then becomes trivial (Pythagoras' Theorem).
A: Let $A$ and $B$ be the tangency points of semicircles $P$ and $Q$ respectively, and let $O$ be the centre of the large semicircle (with radius 10). Then let $R$ be the tangency point of semicircle $Q$ to $O$. 
Extend $LM$ and $QR$ to intersect at $T$. Since $\triangle ATP\sim \triangle BTQ, \frac{RT+4}{4}=\frac{RT+14}{6}\implies RT=16.$
Now let $S$ be on $LT$ such that $OS||QB$. Then $\frac{QT}{OT}=\frac{QB}{OS}$, so $\frac{20}{26}=\frac{4}{OS}$ , hence $OS=\frac{26}{5}$.
Since $OL=OM=10$, we can now directly compute the answer using the Pythagorean theorem 
