In the diagram, two circles touch at $P$. Also, $Q P$ and $S U$ are perpendicular diameters of the larger circle that intersect at $O$. Point $V$ is on $Q P$ and $V P$ is a diameter of the smaller circle. The smaller circle intersects $S U$ at $T$, as shown. If $Q V=9$ and $S T=5$, what is the sum of the lengths of the diameters of the two circles?
(A) 50
(B) 91
(C) 41
(D) 82
(E) 100

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-029.jpg?height=472&width=464&top_left_y=233&top_left_x=1294)
Let $D$ be the length of the diameter of the larger circle and let $d$ be the length of the diameter of the smaller circle.

Since $Q P$ and $V P$ are diameters of the larger and smaller circles, then $Q V=Q P-V P=D-d$. Since $Q V=9$, then $D-d=9$.

Let $C$ be the centre of the smaller circle and join $C$ to $T$. Since $D>d, C$ is to the right of $O$ along $Q P$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-057.jpg?height=483&width=477&top_left_y=504&top_left_x=886)

Since $C T$ is a radius of the smaller circle, then $C T=\frac{1}{2} d$.

Also, $O C=O P-C P$. Since $O P$ and $C P$ are radii of the two circles, then $O C=\frac{1}{2} D-\frac{1}{2} d$.

Since $S O$ is a radius of the larger circle and $S T=5$, then $T O=S O-S T=\frac{1}{2} D-5$.

Since $Q P$ and $S U$ are perpendicular, then $\triangle T O C$ is right-angled at $O$.

By the Pythagorean Theorem,

$$
\begin{aligned}
T O^{2}+O C^{2} & =C T^{2} \\
\left(\frac{1}{2} D-5\right)^{2}+\left(\frac{1}{2} D-\frac{1}{2} d\right)^{2} & =\left(\frac{1}{2} d\right)^{2} \\
4\left(\frac{1}{2} D-5\right)^{2}+4\left(\frac{1}{2} D-\frac{1}{2} d\right)^{2} & =4\left(\frac{1}{2} d\right)^{2} \\
(D-10)^{2}+(D-d)^{2} & =d^{2} \\
(D-10)^{2}+9^{2} & =d^{2} \\
81 & =d^{2}-(D-10)^{2} \\
81 & =(d-(D-10))(d+(D-10)) \\
81 & =(d-D+10)(d+(D-10)) \\
81 & =(10-(D-d))(d+D-10) \\
81 & =(10-9)(d+D-10) \\
81 & =d+D-10 \\
91 & =d+D
\end{aligned}
$$

and so the sum of the diameters is 91 .