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Let the circle be $x^2 + y^2 = r^2$ as shown, where $r$ is its radius. Let the coordinates of $B$ be $(a,b)$ . Then by Pythagorean theorem:

$\begin{cases} a^2 + b^2 = r^2 & ...(1) \\ (a+16)^2 + b^2 = r^2 & ...(2) \\ (a-9)^2 + (b+15)^2 = r^2 & ...(3) \end{cases}$

$\begin{aligned} (2) - (1): \quad 32a + 16^2 & = 0 \\ \implies a & = - 8 \end{aligned}$

$\begin{aligned} (3) - (1): \quad -18{\color{#3D99F6}a} + 81 + 30b + 225 & = 0 & \small \color{#3D99F6} \text{Note that }a = -8 \\ \implies b & = - 15 \end{aligned}$

From $(1): \ r = \sqrt{a^2 + b^2} = \sqrt{(-8)^2+(-15)^2} = \boxed {17}$ .

From the triangle ABC ( Right angled at B) , BC = 15 + 15 = 30, AC is a diameter of the circle,

{AC}^2 = {30}^2 + {16}^2 = 1156

AC = 34

D = 34

R = 17