Cocentric Circles and Probability

For simplicity since the triangle can rotate along the circles, we can consider the version of the triangle with $C$ on the $y$ -axis:

With:

$\large \displaystyle x_C = 0, y_C = 5$

$\large \displaystyle x_A = - x_B, y_A = y_B$

Also, since $A$ and $B$ lie on $X$ :

$\large \displaystyle x_A^2 + y_A^2 = x_B^2 + y_B ^2 = 16$

So, we can say that $\overline{AB} = \overline{AC}$ :

$\large \displaystyle (x_A - x_B)^2 + (y_A - y_B)^2 = (x_A - 0)^2 + (y_A - 5)^2$

$\large \displaystyle x_A^2 + y_A^2 + x_B^2 + y_B^2 - 2x_Ax_B - 2y_Ay_B = x_A^2 + y_A^2 - 10 y_A +25$

Since $x_A^2 + y_A^2 = x_B^2 + y_B ^2 = 16$ :

$\large \displaystyle 32 - 2x_Ax_B - 2y_Ay_B = 16 - 10 y_A +25$

Since $x_A = - x_B, y_A = y_B$ :

$\large \displaystyle 2x_A^2 - 2y_A^2 = -10 y_A + 9$

Since $x_A^2 = 16 - y_A^2$ :

$\large \displaystyle 32 - 4y_A^2 = -10 y_A + 9$

$\large \displaystyle 4y_A^2 -10 y_A - 23 = 0$

So:

$\color{#20A900} \boxed { \large \displaystyle y_A = \frac{5 \pm 3\sqrt{13}}{4} }$

Which leads to:

$\color{#20A900} \boxed { \large \displaystyle x_A = - \sqrt{\frac{114 \pm 30\sqrt{13}}{16} } }$

Each possible side will be $l = x_B - x_A = - 2x_A$ , for each different $x_A$ :

$\large \displaystyle l_1 = 2\sqrt{\frac{114 + 30\sqrt{13}}{16} }$

$\large \displaystyle l_2 = 2\sqrt{\frac{114 - 30\sqrt{13}}{16} }$

$r$ will be equal to $l_1 \cdot l_2$ :

$\color{#20A900} \boxed {\large \displaystyle r = 9 }$

So, the probability will be the area outside $Y$ but within $Z$ divided by the area of the whole $Z$ :

$\large \displaystyle p = \frac{\pi 9^2 - \pi 5^2}{\pi 9^2 }$

$\color{#3D99F6} \boxed{ \large \displaystyle p = \frac{56}{81} }$