If the graph of x^2+y^2=m is tangent to that of x+y=√2m, then:

(E) m may be any nonnegative real number

1969 AMC 12 Problem 21

Attempt Problem 21 of the 1969 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1969 AMC 12 solutions, or check the answer key.

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21.

If the graph of

x^2+y^2=m

is tangent to that of

x+y=\sqrt{2m},

then:

m

must equal

\dfrac12

m

must equal

\dfrac1{\sqrt2}

m

must equal

\sqrt2

m

must equal

2

m

may be any nonnegative real number

The circle has center the origin and radius

\sqrt m

Find the distance from the origin to

x+y-\sqrt{2m}=0

The circle has radius

\sqrt m.

The distance from the origin to the line is

\frac{\lvert-\sqrt{2m}\rvert}{\sqrt{1^2+1^2}}=\sqrt m.

Thus the line is tangent for every

m\ge0

(with the

m=0

case degenerate at the origin).

Therefore, the correct answer is E.