In the diagram, the line segment with endpoints $P(-4,0)$ and $Q(16,0)$ is the diameter of a semi-circle. If the point $R(0, t)$ is on the circle with $t>0$, then $t$ is
(A) 6
(B) 10
(C) 8
(D) 9
(E) 7

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-051.jpg?height=387&width=574&top_left_y=2208&top_left_x=1163)
Since $P(-4,0)$ and $Q(16,0)$ are endpoints of a diameter of the semi-circle, then the length of the diameter is $16-(-4)=20$.

Since a diameter of the semi-circle has length 20 , then the radius of the semi-circle is $\frac{1}{2}(20)=10$. Also, the centre $C$ is the midpoint of diameter $P Q$ and so has coordinates $\left(\frac{1}{2}(-4+16), \frac{1}{2}(0+0)\right)$ or $(6,0)$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-101.jpg?height=385&width=572&top_left_y=1572&top_left_x=836)

Now the distance between $C(6,0)$ and $R(0, t)$ is 10 , since $C R$ is a radius.

Therefore,

$$
\begin{aligned}
\sqrt{(6-0)^{2}+(0-t)^{2}} & =10 \\
36+t^{2} & =100 \\
t^{2} & =64
\end{aligned}
$$

(Alternatively, we could have noted that if $O$ is the origin, then $\triangle R O C$ is right-angled with $R O=t, R C=10$ and $O C=6$ and then used the Pythagorean Theorem to obtain $t^{2}+6^{2}=10^{2}$, which gives $t^{2}=64$.)

Since $t>0$, then $t=8$.