The point $(x,y)$ in the coordinate has a distance of $6$ units from the $x$-axis, a distance of $15$ units from the point $(5,7)$, and a distance of $\sqrt{n}$ from the origin. If both $x$ and $y$ are negative, what is $n$?

Answer 1

We can see from the diagram below that our point must lie in the third quadrant.

[asy]
unitsize (2 cm);

draw((-1,0)--(16,0));
draw((0,-1)--(0,9));
draw(arc((5,7),6,sqrt(130),360),red);

dot((5,7));
label("(5,7)",(5,7),N);

dot((12.810,0));
label("($x,y$)",(12.810,0),NE);
[/asy]

The horizontal distance from $(x,y)$ to the $y$-axis is $x$, so the horizontal distance from $(x,y)$ to the vertical line passing through $(5,7)$ is $x - 5$. Similarly, the vertical distance from $(x,y)$ to the horizontal line passing through $(5,7)$ is $y - 7$. Therefore,
\begin{align*}
(x - 5)^2 + (y - 7)^2 &= 15^2, \\
x^2 + y^2 &= (\sqrt{n})^2 = n.
\end{align*}Therefore, $(x - 5)^2 + (y - 7)^2 + 2(x - 5)(y - 7) + x^2 + y^2 = 15^2 + n.$ Simplifying the left side gives $2x^2 - 24x + 2y^2 - 28y + 50 = 225 + n$. By the first equation, $y^2 = 36$, so
\[2x^2 - 24x + 72 - 28y + 50 = 225 + n.\]Also, $x$ and $y$ are negative, and $y^2 = 36$ implies that $y = -6$. Substituting, we get
\[2x^2 - 24x - 136 = 225 + n.\]Since $225 + n$ must be close to a multiple of 2, and $225 + n$ must be a perfect square, $225 + n$ must be 288. Therefore, $n = \boxed{63}.$