In the diagram, the two circles are centred at $O$. Point $S$ is on the larger circle. Point $Q$ is the point of intersection of $O S$ and the smaller circle. Line segment $P R$ is a chord of the larger circle and touches (that is, is tangent to) the smaller circle at $Q$. Note that $O S$ is the perpendicular bisector of $P R$. If $P R=12$ and $Q S=4$, then the radius of the larger circle is
(A) 6.0
(B) 5.0
(C) 6.5
(D) 7.2
(E) 20.0

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-076.jpg?height=366&width=353&top_left_y=1736&top_left_x=1298)
Suppose that the radius of the larger circle is $r$.

Join $O$ to $P$. Then $O P=O S=r$.

Since $Q$ is the midpoint of $P R$, and $P R=12$, then $P Q=\frac{1}{2} P R=6$.

Since $O S=r$ and $Q S=4$, then $O Q=O S-Q S=r-4$.

Since $\triangle O P Q$ is right-angled at $Q$, then by the Pythagorean Theorem,

$$
\begin{aligned}
O Q^{2}+P Q^{2} & =O P^{2} \\
(r-4)^{2}+6^{2} & =r^{2} \\
r^{2}-8 r+16+36 & =r^{2} \\
52 & =8 r \\
r & =\frac{52}{8}
\end{aligned}
$$

Therefore, the radius of the larger circle is $\frac{52}{8}$, or 6.5 .

ANswer: (C)