Maths Olympiad Prep

Library / /282 of 520

Geometry Difficulty 3.1 AMC 10/12 Find the answer

If rr is positive and the line whose equation is x+y=rx + y = r is tangent to the circle whose equation is x2+y2=rx^2 + y ^2 = r, then rr equals

Pick one

Solution

The circle x2+y2=rx^2 + y^2 = r has center (0,0)(0,0) and radius r\sqrt{r}. Therefore, if the line x+y=rx + y = r is tangent to the circle x2+y2=rx^2 + y^2 = r, then the distance between (0,0)(0,0) and the line x+y=rx + y = r is r\sqrt{r}.
The distance between (0,0)(0,0) and the line x+y=rx + y = r is
0+0r12+12=r2.\frac{|0 + 0 - r|}{\sqrt{1^2 + 1^2}} = \frac{r}{\sqrt{2}}.
Hence,
r2=r.\frac{r}{\sqrt{2}} = \sqrt{r}.
Then r=r2r = \sqrt{r} \cdot \sqrt{2}, so r=2\sqrt{r} = \sqrt{2}, which means r=2r = \boxed{2} or (B), 22.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.