Two Tangent Lines

Since $AC=13$ (By Pythagorean theorem) , let $AP=x$ , therefore $PC=x+13=QC=QB+12$ which implies $QB=x+1$ .

Now let the point of contact of $AB$ with the circle be equal to $O$ . Now we know that $AP=AO=x$ and $BQ=BO=x+1$ , which implies $x+x+1=5 \implies x=2=AP$ . Therefore $BQ=3$ .

Now draw a parallel line through $A$ to $PQ$ to cut $CQ$ at $D$ . Now we can see that $ADQP$ is an isosceles trapezium as $\angle CQP= \angle CPQ$ . This implies $AP=DQ=2$ . So $BD=1$ . Now again by Pythagorean theorem, we have $AD=\sqrt{1^2+5^2}=\sqrt{26}$ .

Since $\triangle CQP$ is similar to $\triangle CDA$ , we have $\dfrac{AD}{PQ}=\dfrac{CD}{CQ}$ (because sides of similar triangles are proportional) $\implies \dfrac{\sqrt{26}}{PQ}=\dfrac{13}{15}$ . Therefore $PQ=\dfrac{15\sqrt{26}}{13}$ making the answer $\boxed{54}$ .

First, $\bar { AC } =\sqrt {5^2+12^2}=13$ .

$\triangle CPQ$ is an isosceles triangle and $\bar { AB}$ is another tangent line, so we get 2 equations.

$\begin{cases} \bar { AP } +\bar { AC } =\bar { BQ } +\bar { BC } \\ \bar {AP}+\bar {BQ}=\bar {AB} \end{cases}$

$\rightarrow \begin{cases} \bar { AP } +13=\bar { BQ } +12\rightarrow \bar { BQ }=\bar {AP }+1 \\ \bar { AP } +\bar { BQ } =5 \end{cases}$

$\rightarrow \begin{cases} \bar { AP } =2 \\ \bar { BQ } =3 \end{cases}$

According to Cosine Rule, we get

$cos~ACB=\frac { 12 }{ 13 } =\frac { 15^{ 2 }+15^{ 2 }-\bar { PQ } ^{ 2 } }{ 2\times15\times15}$

Therefore, $\bar { PQ } =\sqrt { \frac { 450 }{ 13 } } =\frac { 15\sqrt { 26 } }{ 13 }$

So the final answer is $15+26+13=\boxed{54}$ .

Let the center of the circle be $O$ and its radius $r$ . Then we note that $OPC$ is a right triangle with $\angle OCP = \frac 12 \angle C$ . Since $\tan C = \frac 5{12} \implies \tan \angle OCP = \frac 15$ . Now $CP = OP\cot \angle OCP = 5r$ .

Let $AB$ be tangent to the circle at $D$ . Then $OD$ is perpendicular to $AB$ and is parallel to CQ. Hence $\angle DOP = \angle A = \theta$ . Since $\triangle AOD$ and $\triangle AOP$ are congruent, $\angle AOD=\angle AOP = \frac \theta 2$ . Then we have:

$\begin{aligned} \tan \angle AOP & = \frac {AP}{OP} = \frac {CP-\blue{AC}}{OP} & \small \blue{\text{By Pythagorean theorem }AC = 13} \\ \red{\tan \frac \theta 2} & = \frac {5r - \blue{13}}r & \small \red{\text{Since }\tan \theta = \frac {12}5 \implies \tan \frac \theta 2 = \frac 23} \\ \red{\frac 23} & = \frac {5r-13}r \\ \implies r & = 3 \end{aligned}$

Therefore $CP=CQ = 5r = 15$ and $PQ = 2CP \sin \dfrac C2 = \dfrac {30}{\sqrt{26}} = \dfrac {15\sqrt{26}}{13}$ . Hence $a+b+c = 15+26+13 = \boxed{54}$ .

AC = 13 by Pythagoras Theorem in triangle ABC

as they are tangents then BQ=BD and AP=AD let BQ=BD=x and AP=AD=y

also CQ = CP (they are also tangents from a single point)

from the figure we see x+y = 5

and 12+x = 13+y (CQ=CP) this gives us : x-y = 1

so , x+y= 5 and x-y=1

from here we get x=3 and y = 2

hence , CQ=CP=15

now again in Triangle ABC $\cos \Theta$ = $\frac{12}{13}$

And in Triangle PQC we apply cosine formula

so, $\cos \Theta$ = $\frac{225+225-PQ^{2}}{2*15*15}$ = $\frac{12}{13}$

now we can solve and find PQ

Radius of circle can be shown equal to 3, QC =PC=15 and PQ=30/(26)^0.5