# Thread: fun circle challenge problem

1. ## fun circle challenge problem

found this problem today and I've been struggling to wrap my head around it. I find it really interesting and if someone would like to give their two cents, I invite you to do so.
heres the problem:

Two circles of radius R are tangent to each other. A line is drawn tangent to both circles, externally. A third circle is drawn so that it is tangent to the two circles as well as the tangent line. The radius of this smaller circle is r. What is the ratio of the circumfrances and the area's for the big to small circles?

2. ## Re: fun circle challenge problem

Is this a trick question? R/r and (R/r)^2?

3. ## Re: fun circle challenge problem

Here's how I would do the problem. Set up a coordinate system so that the center of one of the first two circles is at (0, 0) and the center of the other is at (2R, 0). The lines y= R and y= -R are tangent to the two circles. The two circles touch at (R, 0) and the small circle would touch y= R at (R, R). Now a line through (0, 0) with slope m has equation y= mx. That will cross x= R at (R, mR) so that the distance from that point to y= R is r= R- mR, and that will be the radius of the small circle. It will cross the circle $x^2+ y^2= R^2$ where $x^2+ m^2x^2= (1+ m^2)x^2= R^2$ or $(\frac{R}{\sqrt{1+m^2}}, \frac{mR}{\sqrt{1+ m^2}})$. The distance from that point to (R, mR) is $R(\sqrt{1+ m^2}- 1)$. Solve $R(\sqrt{1+ m^2}- 1)= R- mR$ for m (obviously the "R"s cancel): $\sqrt{1+ m^2}= 2- m$. Squaring both sides, $1+ m^2= 4- 4m+ m^2$ so m= 3/4 and then r= R- (3/4)R= (1/4)R.

4. ## Re: fun circle challenge problem

any other ideas for ways to solve it?

5. ## Re: fun circle challenge problem

I drew up a quick diagram that contains a right triangle. im not sure what to do now though.

6. ## Re: fun circle challenge problem

Ok i see it now. Use the pythagorean theorem, based on that picture: $(R-r)^2 + R^2 = (R+r)^2$
$R^2-2rR+r^2+R^2=R^2+2rR+r^2$
$R^2=4rR$
$R = 4r$

cool. thanks