Let $P = (x,y)$ be a point in the coordinate plane. The point $P$ has a distance of $12$ units from the $x$-axis, and it has a distance of $10$ units from the point $(2,5)$. If $P$ has a distance of $n$ units from the origin, and $x > 1$, what is $n$?
Since $P$ is $12$ units from the $x$-axis, $(x,y)$ lies on a line parallel to the $y$-axis, $12$ units away. So, we have \[|y| = 12. \quad (*)\]Since $P$ is $10$ units from $(2,5)$ and $x > 1$, $(x,y)$ lies on a circle with center $(2,5)$ and radius $10$. In other words, \[\sqrt{(x - 2)^2 + (y - 5)^2} = 10. \quad (**) \sqrt{(x - 2)^2 + (y + 5)^2} = 10. \quad (***) \]To find $n$, we want to compute the distance from $(x,y)$ to the origin. This is $\sqrt{x^2 + y^2}$. We can use $(*)$ and $(**)$ to eliminate $y$. From $(*)$, $y = 12$ or $y = -12$. Substituting into $(**)$, we get $\sqrt{(x - 2)^2 + 7^2} = 10$ and $\sqrt{(x - 2)^2 + (-7)^2} = 10$. Squaring both sides of both equations, we get $(x - 2)^2 + 7^2 = 100$ and $(x - 2)^2 + (-7)^2 = 100$. Solving these equations, we get $(x - 2)^2 = 51$ and $(x - 2)^2 = 51$. Thus, $x = \pm 7$. If $x = 7$, from $(*)$ we have $y = -12$, so $\sqrt{x^2 + y^2} = \sqrt{7^2 + (-12)^2} = \sqrt{49 + 144} = \sqrt{193}$. If $x = -7$, from $(*)$ we have $y = 12$, so $\sqrt{x^2 + y^2} = \sqrt{(-7)^2 + 12^2} = \sqrt{49 + 144} = \sqrt{193}$. So, from $(*)$ and $(**)$, $n = \boxed{\sqrt{193}}$.
To solve this problem, we can use the distance formula. The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula:
$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
In this case, we know that the point $P = (x, y)$ has a distance of $12$ units from the $x$-axis. Since the $x$-axis is a horizontal line with equation $y = 0$, we can substitute $y_2 = 0$ into the distance formula:
$$12 = \sqrt{(x - x_1)^2 + (0 - y_1)^2}$$
Simplifying this equation, we get:
$$144 = (x - x_1)^2 + (0 - y_1)^2$$
Since $y_1 = 0$, this equation becomes:
$$144 = (x - x_1)^2$$
Since $x_1 = 2$ (the $x$-coordinate of the point $(2, 5)$), we have:
$$144 = (x - 2)^2$$
Taking the square root of both sides, we get:
$$12 = x - 2$$
Solving for $x$, we find that $x = 14$.
Now, we can use the distance formula to find the distance between the points $P = (14, y)$ and $(2, 5)$:
$$10 = \sqrt{(14 - 2)^2 + (y - 5)^2}$$
Simplifying this equation, we get:
$$100 = (12)^2 + (y - 5)^2$$
$$100 = 144 + (y - 5)^2$$
$$-44 = (y - 5)^2$$
Since $y > 1$, the equation $(y - 5)^2 = -44$ has no real solutions.
Therefore, there is no point $P$ that satisfies all the given conditions.