2011-I-16

The mid-point of $PR$

$\begin{array}{cl}
= & \left( \dfrac{16+64}{2}, \dfrac{80+(-48)}{2} \right) \\
= & (40,16)
\end{array}$

The slope of $PR$

$\begin{array}{cl}
= & \dfrac{80 – (-48)}{16-64} \\
= & \dfrac{-8}{3}
\end{array}$

Therefore, the slope of the perpendicular bisector

$\begin{array}{cl}
= & -1 \div \dfrac{-8}{3} \\
= & \dfrac{3}{8}
\end{array}$

Hence, the equation of the perpendicular bisector of $PR$ is

$\begin{array}{rcl}
y-16 & = & \dfrac{3}{8}(x-40) \\
8y – 128 & = & 3x – 120 \\
3x-8y + 8 & = & 0
\end{array}$

Sub. $x=16$ into $3x-8y+8=0$, we have

$\begin{array}{rcl}
3(16) – 8y +8 & = & 0 \\
8y & = & 56 \\
y & = & 7
\end{array}$

Therefore, the coordinates of the circumcentre of $\Delta PQR$ are $(16,7)$.

1. The radius of the circle

   $\begin{array}{cl}
   = & 80 – 7 \\
   = & 73
   \end{array}$

   Therefore, the equation of $C$ is $(x-16)^2+(y-7)^2=73^2$.

2. Since $\Delta PQR$ is an isosceles triangle with $PQ=PR$, then $PS\perp QR$, i.e. $\angle PSQ=90^\circ$. Let $K$ be the centre of $C$. If $K$ is also the in-centre, then $KQ$ is the angle bisector of $\angle PQS$.

   In $\Delta PSQ$,

   $\begin{array}{rcl}
   \tan \angle PQS & = & \dfrac{PS}{QS} \\
   & = & \dfrac{80-(-48)}{16 – (-32)} \\
   \angle PQS & = & 69.443~954~78^\circ
   \end{array}$

   In $\Delta KSQ$,

   $\begin{array}{rcl}
   \tan \angle KQS & = & \dfrac{KS}{QS} \\
   & = & \dfrac{7-(-48)}{16-(-32)} \\
   \angle KQS & = & 48.887~909~56^\circ
   \end{array}$

   Since $\angle KQS \neq \dfrac{1}{2}\angle PQS$, then $KQ$ is not the angle bisector of $\angle PQR$. Therefore, the centre of $C$ and the in-centre of $\Delta PQR$ are not the same point.