2019-I-19

Therefore, the graph of $y=f(x)$ passes through $F$.

1. $\begin{array}{rcl}
   g(x) & = & f(-x) + 4 \\
   & = & \dfrac{1}{1+k}[(-x)^2 + (6k -2)(-x) +(9k +25)] + 4 \\
   & = & \dfrac{1}{1 + k}[x^2 – (6k – 2)x + (\dfrac{6k-2}{2})^2 – (\dfrac{6k-2}{2})^2 + (9k +25)] + 4 \\
   & = & \dfrac{1}{1 + k}[(x – (3k – 1))^2 – (3k -1)^2 + (9k + 25)] + 4 \\
   & = & \dfrac{1}{1+k} [x – (3k – 1)]^2 + \dfrac{-(3k – 1)^2 +(9k+25)+ 4(1+k)}{1+k} \\
   & = & \dfrac{1}{1+k}[x – (3k – 1)]^2 + \dfrac{-9k^2 + 6k – 1 + 9k + 25 + 4 + 4k}{1+k} \\
   & = & \dfrac{1}{1+k}[x – (3k – 1)]^2 + \dfrac{-9k^2 + 19k + 28}{1+k} \\
   & = & \dfrac{1}{1+k}[x – (3k – 1)]^2 + \dfrac{(28-9k)(1+k)}{1+k} \\
   & = & \dfrac{1}{1+k}[x – (3k – 1)]^2 + (28 – 9k)
   \end{array}$

   Therefore, the coordinates of $U$ are $(3k-1, 28-9k)$.

2. For the area of the circle passing through $F$, $O$ and $U$ is the least, then $FO$ must be a diameter of the circle.Therefore, $\angle FUO = 90^\circ$ ($\angle$ in semi-circle).

   For $\angle FUO = 90^\circ$, we have

   $\begin{array}{rcl}
   m_{FU} \times m_{UO} & = & -1 \\
   \dfrac{33 – (28 – 9k)}{4 – (3k-1)} \times \dfrac{(28 – 9k) – 0}{(3k – 1) – 0} & = & -1 \\
   (5 + 9k)(28 – 9k) & = & -(5-3k)(3k – 1) \\
   140 + 207k – 81 k^2 & = & 9k^2 -18k + 5 \\
   90k^2 – 225k – 135 & = & 0 \\
   2k^2 -5k -3 & = & 0 \\
   (k – 3)(2k + 1) & = & 0
   \end{array}$

   Therefore, $k=3$ or $k = \dfrac{-1}{2}$ (rejected).

3. Note that for any positive value of $k$, $y=f(x)$ passes through $F(4,33)$ and $y=g(x)$ passes through $G$. Hence, $G$ is the image of reflecting $F$ with respect to the $y$-axis and then translating upwards by $4$ units. Then, we have$\begin{array}{rcl}
   G & = & (-4, 33 + 4) \\
   G & = & (-4, 37)
   \end{array}$

   For the area of the circle passing through $F$, $O$ and $V$ is the least, $FO$ must be a diameter. Then $\angle FVO = 90^\circ$ ($\angle$ in semi-circle).

   Consider $\angle FGO$.

   $\begin{array}{cl}
   & m_{FG} \times m_{GO} \\
   = & \dfrac{33 – 37}{4 -(- 4)} \times \dfrac{37 – 0}{-4 – 0} \\
   = & \dfrac{37}{8} \\
   \neq & -1
   \end{array}$

   Therefore, $\angle FGO \neq 90^\circ$.

   Hence, $\angle FVO + \angle FGO \neq 180^\circ$.

   Therefore, $F$, $G$, $O$ and $V$ are not concyclic.