2014-I-12

$\begin{array}{cl}
= & \sqrt{(6-0)^2 + (11-3)^2} \\
= & 10
\end{array}$

Hence, the equation of $C$ is

$\begin{array}{rcl}
(x-0)^2 + (y-3)^2 & = & 10^2 \\
x^2 +y^2 – 6y – 91 & = & 0
\end{array}$

1. Let $P=(x,y)$.

   $\begin{array}{rcl}
   AP & = & GP \\
   \sqrt{(x-6)^2+(y-11)^2} & = & \sqrt{(x-0)^2+(y-3)^2} \\
   x^2 -12x +36 +y^2 -22y+121 & = & x^2 + y^2 -6y +9 \\
   -12x -16y + 148 & = & 0 \\
   3x+4y-37 & = & 0
   \end{array}$

2. $\Gamma$ is the perpendicular bisector of $AG$.
3. Let $T$ be the intersection point of $GA$ and $QR$. Note that $GA\perp QR$. Hence $QT=TR$. Also, since $\Gamma$ is the perpendicular bisector of $AG$, then $AT=TG$. Therefore, quadrilateral $AQGR$ is a rhombus. Hence, the perimeter of the quadrilateral $AQGR$

   $\begin{array}{cl}
   = & 4 \times AQ \\
   = & 4 \times \text{ the radius of $C$} \\
   = & 4 \times 10 \\
   = & 40
   \end{array}$