2015-I-14

1. The mid-point of $PQ$

   $\begin{array}{cl}
   = & \left( \dfrac{4+(-14)}{2}, \dfrac{-1+23}{2} \right) \\
   = & ( -5, 11)
   \end{array}$

   The slope of $PQ$

   $\begin{array}{cl}
   = & \dfrac{23-(-1)}{-14 – 4} \\
   = & \dfrac{-4}{3}
   \end{array}$

   Since $L\perp PQ$, then the slope of $L$

   $\begin{array}{cl}
   = & -1 \div \dfrac{-4}{3} \\
   = & \dfrac{3}{4}
   \end{array}$

   Hence, the equation of $L$ is

   $\begin{array}{rcl}
   y-11 & = & \dfrac{3}{4}(x-(-5)) \\
   4y – 44 & = & 3x + 15 \\
   3x – 4y + 59 & = & 0
   \end{array}$

2. Let $G=(h,k)$. Note that $PG$ is a radius of the circle $C$, then the radius

   $\begin{array}{cl}
   = & \sqrt{(4-h)^2+(-1 – k)^2} \\
   = & \sqrt{16 – 8h +h^2 + 1 +2k + k^2} \\
   = & \sqrt{h^2 + k^2 -8h +2k + 17}
   \end{array}$

   Then the equation of the circle $C$ is

   $\begin{array}{rcl}
   (x-h)^2 + (y-k)^2 & = & h^2 + k^2 -8h +2k + 17 \\
   x^2 -2hx +h^2 + y^2 -2ky +k^2 & = & h^2 + k^2 -8h +2k + 17 \\
   x^2 +y^2 -2hx -2ky +8h -2k -17 & = & 0
   \end{array}$

   Since $G$ is lying on $L$, then we have

   $\begin{array}{rcl}
   3h-4k+59 & = & 0 \\
   k & = & \dfrac{3h+59}{4}
   \end{array}$

   Sub. $k=\dfrac{3h+59}{4}$ into the equation of $C$, we have

   $\begin{array}{rcl}
   x^2 + y^2 -2hx – 2\left(\dfrac{3h+59}{4} \right)y +2h – 2\left(\dfrac{3h+59}{4} \right) -17 & = & 0 \\
   2x^2 + 2y^2 -4hx -(3h+59)y +16h – 3h -59 – 34 & = & 0 \\
   2x^2 + 2y^2 -4hx -(3h+59)y +13h – 93 & = & 0
   \end{array}$

   Therefore, the equation of the circle $C$ is $2x^2 + 2y^2 -4hx -(3h+59)y +13h – 93 = 0 $.

$\begin{array}{rcl}
2(26)^2 + 2(43)^2 -4h(26) -(3h+59)(43) +13h – 93 & = & 0 \\
5050 – 104h – 129h -2537 +13h -93 & = & 0 \\
220 & = & 2420 \\
h & = & 11
\end{array}$

Therefore, the equation of the circle passing through $P$, $Q$ and $R$ is

$\begin{array}{rcl}
2x^2 +2y^2 -4(11)x -(3(11)+59)y +13(11)-93 & = & 0 \\
2x^2 +2y^2 -44x – 92 y +50 & = & 0 \\
x^2 + y^2 -22x -46y +25 & = & 0
\end{array}$

Therefore, the radius of the circle

$\begin{array}{cl}
= & \sqrt{\left(\dfrac{-22}{2}\right)^2+ \left(\dfrac{-46}{2} \right)^2 – 25} \\
= & 25
\end{array}$

Hence, the diameter of the circle

$\begin{array}{cl}
= & 2 \times 25 \\
= & 50
\end{array}$