Circle inscribed in three intersecting lines.

Since the result is a unique number, it is independent of $\theta$ , so, we can use any convenient value, like $\theta=0$ for instance. Then, we have the following three lines $\begin{aligned} & {{l}_{1}}:3y-4x=0 \\ & {{l}_{2}}:4x+3y=24 \\ & {{l}_{3}}:y=0 \\ \end{aligned}$ The rest is simple calculations: The intersection points of the lines are easily calculated ${{l}_{1}}\cap {{l}_{2}}\equiv A\left( 3,4 \right) \ \ \ \ \ {{l}_{2}}\cap {{l}_{3}}\equiv B\left( 6,0 \right) \ \ \ \ \ {{l}_{3}}\cap {{l}_{1}}\equiv C\left( 0,0 \right)$ The perimeter of $\triangle ABC$ defined by these points is $P=AB+BC+CA=\sqrt{{{\left( 6-3 \right)}^{2}}+{{\left( 0-4 \right)}^{2}}}+\sqrt{{{\left( 0-6 \right)}^{2}}+{{\left( 0-0 \right)}^{2}}}+\sqrt{{{\left( 3-0 \right)}^{2}}+{{\left( 4-0 \right)}^{2}}}=16$ The area of the triangle is $\left[ ABC \right]=\dfrac{1}{2}BC\cdot d\left( A,BC \right)=\dfrac{1}{2}\times 6\times 4=12$ For the radius of the incircle of $\triangle ABC$ we have $r=\frac{\left[ ABC \right]}{\frac{1}{2}P}=\frac{12}{8}=\frac{3}{2}$ Finally, the area of the incircle is ${{A}_{c}}=\pi {{r}^{2}}=\frac{9}{4}\pi$ For the answer, $a=9$ , $b=4$ , thus, $A+B=\boxed{13}$ .

$(1): \:\ (3\cos(\theta) - 4\sin(\theta))y - (4\cos(\theta) + 3\sin(\theta))x = 0$

$(2): \:\ (4\cos(\theta) - 3\sin(\theta))x + (3\cos(\theta) + 4\sin(\theta))y = 24$

$(3): \:\ (y\cos(\theta) - x\sin(\theta) = 0$

This is a fairly basic problem in which I present two methods. The first method I use a rotation thru an angle $\theta$ . The second method I just solved the equations in pairs to find the points of intersection.

Method 1:

Using the rotation equations

$x = x'\cos(\theta) - y'\sin(\theta)$

$y = y'\sin(\theta) + y'\cos(\theta)$

and replacing the equations above into $(1), (2)$ and $(3)$ we obtain:

$(1'): \:\ y' = \dfrac{4}{3}x'$

$(2'): \:\ y' = -\dfrac{4}{3}(x' - 6)$

$(3'): \:\ y' = 0$

Solving $1'$ and $2' \implies x' = y' = 0 \implies A:(0,0)$

Solving $2'$ and $3' \implies y' = 0$ and $x' = 6 \implies C:(6,0)$

Solving $1'$ and $2' \implies x' = 3$ and $y' = 4 \implies B:(3,4)$

$\implies \overline{AB} = \overline{BC} = 5$ and $\overline{AC} = 6$

$A_{\triangle{ABC}} = \dfrac{1}{2}(6)(4) = 12 = \dfrac{1}{2}r(16) = 8r \implies r = \dfrac{3}{2} \implies A_{c} = \dfrac{9}{4}\pi = \dfrac{\alpha}{\beta}\pi$

$\implies \alpha + \beta = \boxed{13}$

Method 2:

For $(1)$ and $(3)$ :

$(3) \implies y = x\tan(\theta) \implies x = y = 0 \implies A:(0,0)$

For $(2)$ and $(3)$ :

$(3) \implies y = x\tan(\theta) \implies x = 6\cos(\theta)$ and $y = 6\sin(\theta) \implies$

$C:(6\cos(\theta), 6\sin(\theta)) \implies \overline{AC} = 6$

For $(1)$ and $(2)$ :

$-(3\cos(\theta) + 4\sin(\theta)) * (1) + (3\cos(\theta) - 4\sin(\theta)) * (2) \implies$

$x = 3\cos(\theta) - 4\sin(\theta)$

and

$(4\cos(\theta) - 3\sin(\theta)) * (1) + (4\cos(\theta) + 3\sin(\theta)) * (2) \implies y = 4\cos(\theta) + 3\sin(\theta)$

$\implies B:(3\cos(\theta) - 4\sin(\theta), 4\cos(\theta) + 3\sin(\theta)) \implies$

$\overline{AB} = \overline{BC} = 5$

$D$ is a midpoint of $\overline{AC} \implies D:(3\cos(\theta), 3\sin(\theta))$ and from above

$B:(3\cos(\theta) - 4\sin(\theta), 4\cos(\theta) + 3\sin(\theta)) \implies \overline{BD} = 4$

$\implies$

$A_{\triangle{ABC}} = \dfrac{1}{2}(6)(4) = 12 = \dfrac{1}{2}r(16) = 8r \implies r = \dfrac{3}{2} \implies A_{c} = \dfrac{9}{4}\pi = \dfrac{\alpha}{\beta}\pi$