14. The points $P(d,-d)$ and $Q(12-d, 2 d-6)$ both lie on the circumference of the same circle whose centre is the origin.
What is the sum of the two possible values of $d$ ?
A -16
B -4
C 4
D 8
E 16
Solution $\square$
Let $O(0,0)$ be the origin.
Because the points $P$ and $Q$ lie on the same circle with centre $O$, we have $O P^{2}=O Q^{2}$. That is,
$$
d^{2}+(-d)^{2}=(12-d)^{2}+(2 d-6)^{2} .
$$

Expanding both sides of this equation, we obtain
$$
d^{2}+d^{2}=\left(144-24 d+d^{2}\right)+\left(4 d^{2}-24 d+36\right) .
$$

We can rearrange this equation to obtain
$$
3 d^{2}-48 d+180=0 .
$$

By dividing both sides of this last equation by 3 , it follows that
$$
d^{2}-16 d+60=0 \text {. }
$$

We can now use the fact that the sum of the roots of the quadratic equation $x^{2}+p x+q=0$ is $-p$ to deduce that the sum of the two possible values of $d$ is $-(-16)$, that is, 16 .