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 $n$?

Answer 1

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 $(**)$, $n = \boxed{\sqrt{193}}$.