C1. $A B C$ is a triangle with coordinates $A=(2,6), B=(0,0)$, and $C=(14,0)$.
(a) Let $P$ be the midpoint of $A B$. Determine the equation of the line perpendicular to $A B$ passing through $P$.
(b) Let $Q$ be the point on line $B C$ for which $P Q$ is perpendicular to $A B$. Determine the length of $A Q$.
(c) There is a (unique) circle passing through the points $A, B$, and $C$. Determine the radius of this circle.
Solution:
(a) The answer is $y=-1 / 3 \cdot x+10 / 3$ or $x+3 y=10$.

The midpoint of $A B$ is
$$
P=\left(\frac{0+2}{2}, \frac{0+6}{2}\right)=(1,3) .
$$

The slope of $A B$ is $6 / 2=3$. Therefore, the slope of the line perpendicular to $A B$ is $-1 / 3$. Hence, the equation of the line perpendicular to $A B$ passing through $P$ is
$$
y-3=\frac{-1}{3}(x-1) \text {. }
$$

This is equivalent to
$$
y=\frac{-1}{3} x+\frac{10}{3} .
$$

Rewriting this yields
$$
x+3 y=10 \text {. }
$$
(b) The answer is 10 .

Solution 1: The line $B C$ is the line $y=0$. Since $Q$ lies on $B C$, the $y$-coordinate of $Q$ is 0 . Since $Q$ also lies on the line passing through $P$ perpendicular to $A B$ and the equation of this line is $x+3 y=10$, we substitute $y=0$ into $x+3 y=10$ to yield $x=10$. Hence, $Q=(10,0)$. Since $A=(2,6)$, by the Pythagorean Theorem,
$$
A Q=\sqrt{(10-2)^{2}+(0-6)^{2}}=\sqrt{8^{2}+6^{2}}=10 .
$$

Solution 2: As in Solution 1, $Q=(10,0)$. Since $Q$ lies on the perpendicular bisector of $A B$, $Q A=Q B$. Since $Q=(10,0)$ and $B=(0,0), Q A=Q B=10$.
(c) The answer is $5 \sqrt{2}$ or $\sqrt{50}$.

Solution 1: Let $O=(x, y)$ be the centre of the circle. Since $O$ lies on the perpendicular bisector of $B C, x=(0+14) / 2=7$. Since $O$ lies on the line perpendicular to $A B$ passing through $P$ and the equation of the line passing through $P$ perpendicular to $A B$ is $x+3 y=10$, we substitute $x=7$ into $x+3 y=10$ to yield $y=1$. Hence, the centre of the circle is at $(7,1)$. The radius of the circle is the distance from $O$ to any of $A, B, C$. For simplicity's sake, we compute the length of $O B$, since $B=(0,0)$. By the Pythagorean Theorem, the radius of the circle is $O B=\sqrt{7^{2}+1^{2}}=\sqrt{50}=5 \sqrt{2}$.

Solution 2: We will use the following property of a triangle; let $a, b, c$ be the side lengths of a triangle, $R$ the circumradius of the triangle and $K$ the area of the triangle. Then the quantities $a, b, c, R, K$ have the following relationship;
$$
K=\frac{a b c}{4 R} .
$$

In this triangle, $A B=\sqrt{2^{2}+6^{2}}=\sqrt{40}=2 \sqrt{10}, B C=14$ and $C A=\sqrt{(14-2)^{2}+6^{2}}=$ $\sqrt{180}=3 \sqrt{20}$. Note that
$$
K=\frac{1}{2} \times B C \times\{\text { the height to side } \mathrm{BC}\}=\frac{1}{2} \times 14 \times 6=42 \text {. }
$$

Therefore,
$$
R=\frac{A B \cdot B C \cdot C A}{4 K}=\frac{2 \sqrt{10} \times 14 \times 3 \sqrt{20}}{4 \times 42}=\frac{60 \sqrt{2} \times 14}{4 \times 42}=5 \sqrt{2} .
$$