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 $n$?
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, $n = \boxed{63}.$
To find the value of $n$, we need to determine the distance between the origin and the point $(x, y)$. We can use the distance formula to calculate this distance.
The distance formula is given by:
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
In this case, the coordinates of the origin are $(0,0)$, and the coordinates of the point are $(x, y)$. Plugging these values into the distance formula gives:
$d = \sqrt{(0 - x)^2 + (0 - y)^2}$
Since $x$ and $y$ are both negative, the expression simplifies to:
$d = \sqrt{x^2 + y^2}$
We are given that the distance between the point $(x, y)$ and the origin is $\sqrt{n}$. Thus, we have:
$\sqrt{n} = \sqrt{x^2 + y^2}$
Squaring both sides of the equation, we get:
$n = x^2 + y^2$
To find the values of $x$ and $y$ that satisfy the distances given in the problem, we can set up a system of equations using the given information:
$x^2 + (6)^2 = 15^2$
$x^2 + 36 = 225$
$x^2 = 189$
$x = \pm \sqrt{189}$
Since $x$ is negative, we can take the negative square root:
$x = -\sqrt{189}$
Similarly, we can find the value of $y$ using the distance formula between $(x, y)$ and $(5, 7)$:
$d = \sqrt{(x - 5)^2 + (y - 7)^2}$
$15 = \sqrt{(\sqrt{189} - 5)^2 + (y - 7)^2}$
Squaring both sides and simplifying, we get:
$225 = (\sqrt{189} - 5)^2 + (y - 7)^2$
$225 = 189 - 10\sqrt{189} + 25 + y^2 - 14y + 49$
$0 = y^2 - 14y + 10\sqrt{189} -11$
Solving this equation for $y$ gives two possible values:
$y = \frac{14 \pm \sqrt{196 - 4(1)(10\sqrt{189} -11)}}{2}$
$y = \frac{14 \pm \sqrt{196 -40\sqrt{189} +44}}{2}$
$y = 7 \pm \sqrt{(6 - \sqrt{189})^2}$
Since $y$ is negative, we take the negative square root:
$y = 7 - \sqrt{(6 - \sqrt{189})^2}$
Therefore, the values of $x$ and $y$, when both are negative, are:
$x = -\sqrt{189}$
$y = 7 - \sqrt{(6 - \sqrt{189})^2}$
To find $n$, we substitute these values of $x$ and $y$ into the equation $n = x^2 + y^2$:
$n = (-\sqrt{189})^2 + \left(7 - \sqrt{(6 - \sqrt{189})^2}\right)^2$
Simplifying this expression will give us the value of $n$.