$26 \cdot 30$ If the graph of $x^{2}+y^{2}=m$ is tangent to the graph of $x+y=\sqrt{2 m}$, then
(A) $m=\frac{1}{2}$.
(B) $m=\frac{1}{\sqrt{2}}$.
(C) $m=\sqrt{2}$.
(D) $m=2$.
(E) $m$ can be any non-negative real number.
(20th American High School Mathematics Examination, 1969)
[Solution] The distance from any point $\left(x_{1}, y_{1}\right)$ on the plane to the line $x+y=\sqrt{2 m}$ is
$$
d=\left|\frac{x_{1}+y_{1}-\sqrt{2 m}}{\sqrt{2}}\right|=\left|\frac{x_{1}+y_{1}}{\sqrt{2}}-\sqrt{m}\right| \text {, }
$$

Then the distance from the origin $(0,0)$ to the line is $\sqrt{m}$, which means the line is tangent to the circle $x^{2}+y^{2}=m$, and $m$ can be any non-negative real number.
Therefore, the answer is $(E)$.