The distance from point $P(x, y)$ to the origin $O(0,0)$ is 17 .

The distance from point $P(x, y)$ to $A(16,0)$ is also 17 .

What are the two possible pairs of coordinates $(x, y)$ for $P$ ?
Solution 1

Since the distance from $(0,0)$ to $(x, y)$ is 17 , then $x^{2}+y^{2}=17^{2}$.

Since the distance from $(16,0)$ to $(x, y)$ is 17 , then $(x-16)^{2}+y^{2}=17^{2}$.

Subtracting the second of these equations from the first, we obtain $x^{2}-(x-16)^{2}=0$ which gives $x^{2}-\left(x^{2}-32 x+256\right)=0$ and so $32 x=256$ or $x=8$.

Since $x=8$ and $x^{2}+y^{2}=17^{2}$, then $64+y^{2}=289$ which gives $y^{2}=225$, from which we get $y=15$ or $y=-15$.

Therefore, the two possible pairs of coordinates for $P$ are $(8,15)$ and $(8,-15)$.

## Solution 2

The point $P$ is equidistant from $O$ and $A$ since $O P=P A=17$.

Suppose that $M$ is the midpoint of $O A$.

Since $O$ has coordinates $(0,0)$ and $A$ has coordinates $(16,0)$, then $M$ has coordinates $(8,0)$.

Since $O P=P A$, then $\triangle O P A$ is isosceles.

This means that median $P M$ in $\triangle O P A$ is also an altitude; in other words, $P M$ is perpendicular to $O A$.

Since $O A$ is horizontal, $P M$ is vertical, and so $P$ lies on the vertical line with equation $x=8$. Since $O M=8$ and $O P=17$ and $\triangle P M O$ is right-angled at $M$, then by the Pythagorean Theorem, $P M=\sqrt{O P^{2}-O M^{2}}=\sqrt{17^{2}-8^{2}}=\sqrt{225}=15$.

Since $P M$ is vertical and $M$ is on the $x$-axis, then $P$ is a distance of 15 units vertically from the $x$-axis.

Since $P$ has $x$-coordinate 8 and is 15 units away from the $x$-axis, then the two possible pairs of coordinates for $P$ are $(8,15)$ and $(8,-15)$.