# 1969 AHSME Problems/Problem 21

## Problem

If the graph of $x^2+y^2=m$ is tangent to that of $x+y=\sqrt{2m}$, then: $\text{(A) m must equal } \tfrac{1}{2}\quad \text{(B) m must equal } \frac{1}{\sqrt{2}}\quad\\ \text{(C) m must equal } \sqrt{2}\quad \text{(D) m must equal } 2\quad\\ \text{(E) m may be any non-negative real number}$

## Solution 1

Note that the first equation represents a circle and the second equation represents a line. If a line is tangent to a circle, then it only hits at one point, so there will only be one solution to the system of equations.

In the second equation, $y = -x + \sqrt{2m}$. Substitution results in $$x^2 + x^2 - 2x\sqrt{2m} + 2m = m$$ $$2x^2 - 2x\sqrt{2m} + m = 0$$ In order for the system to have one solution, the discriminant must equal $0$. $$4(2m) - 4 \cdot 2 \cdot m = 0$$ $$0 = 0$$ Thus, $m$ can be a non-negative real number, so the answer is $\boxed{\textbf{(E)}}$.

## Solution 2

Since $x^2+y^2=m$ has radius $\sqrt{m}$, and $x+y=\sqrt{2m}$ is a diagonal line in the plane with slope $-1$, if the two curves are tangent then $x+y=\sqrt{2m}$ either has to pass through $(\frac{\sqrt{2m}}{2}, \frac{\sqrt{2m}}{2})$ or $(-\frac{\sqrt{2m}}{2}, -\frac{\sqrt{2m}}{2})$. However, as $\sqrt{2m}$ is positive, the line can only pass through $(\frac{\sqrt{2m}}{2}, \frac{\sqrt{2m}}{2})$, which it does for all $m$. Thus the answer is $\boxed{\textbf{(E)}}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 