\(PB = QB = radius = r\) (Let us say)

\(AP = \frac{1}{2}PB\)
\(AP = \frac{r}{2}\)

\(AB = AP + PB = \frac{r}{2} + r = \frac{3r}{2}\)

\(CQ = \frac{2}{7}QB\)
\(CQ = \frac{2r}{7}\)

\(BC = CQ + QB = \frac{2r}{7} + r = \frac{9r}{7}\)

Since, ABC is a Right angled triangle, we can apply Pythagoras theorem;

\(AC^2 = AB^2 + BC^2\)
\(100^2 = (\frac{3r}{2})^2 + (\frac{9r}{7})^2\)

\(100^2 = \frac{9r^2}{4} + \frac{81r^2}{49}\)

\((4)(49)100^2 = (9r^2)(49) + (81r^2)(4)\)

\((4)(49)100^2 = 765r^2\)

\(r^2 = \frac{(4)(49)100^2}{765}\)

\(r = \sqrt{\frac{(4)(49)100^2}{765}} = \frac{1400}{\sqrt{765}}\)

\(r = \frac{1400}{3\sqrt{85}}\)

Since, we have \(\sqrt{85}\) in option A - it must be the our Answer!