Problem 390

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 $PQ$ and so has coordinates $\left(\frac{1}{2}(-4+16),\frac{1}{2}(0+0)\right)$ or $(6,0)$.

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Now the distance between $C(6,0)$ and $R(0,t)$ is 10, since $CR$ 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 $\triangle ROC$ is right-angled with $RO=t$, $RC=10$ and $OC=6$ and then used the Pythagorean Theorem to obtain $t^2+6^2=10^2$, which gives $t^2=64$.)

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