# Find the radius of the third circle given three circles

Question: A circle with the center at a point $A$ and a radius $r$ touches internally a circle with the center at a point $B$ and a radius $R$. A third circle touches each of the circles and the line $AB$. Prove that the radius of the third circle is equal to $$\frac{4*r*R*(R-r)}{(R+r)^2}$$

I tried to solve the question, but I got confused by the placement of the circles. In addition, I am not sure how this question ties in with triangles (The unit this question is part of is on trignometry). Would I need to construct triangles around the circles in order to solve this question?

• By "touching internally", do you mean they intersect (overlap)? – AspiringMathematician Apr 26 '17 at 1:27
• I'm not sure, as I received this question from my teacher and he did not specify. However, I believe that it does mean that they overlap. – jenboo12138 Apr 26 '17 at 1:35
• Are you using the word "touch" to mean "tangent to"? – John Wayland Bales Apr 26 '17 at 1:37
• I am not sure (these are the exact wording of the problem I received), but I assume that they do mean that they are tangent. – jenboo12138 Apr 26 '17 at 1:45
• The meanings are far too unclear to make any conclusion of any sort. Your teacher should be ashamed of himself and you should start seeking a refund. The question is unanswerable and meaningless. All we can conclude is d (A,B) <= (R+r)/2. The third circle can be any ... well there are a few things it can't be but not many. worthless question. – fleablood Apr 26 '17 at 1:48

I think "two circles touch internally" means they are tangent and one is inside another. You problem settings would look like the following picture.

Now, if we let $\rho$ be the radius of the third circle, then $$AB= R-r, BC=R-\rho, CA=r+\rho,CE=\rho.$$

So, $\triangle ABC$ has half perimeter $$s = \frac{AB+BC+CA}2 = R.$$

By Heron's formula, $$area(\triangle ABC) = \sqrt{Rr\rho(R-r-\rho)}.$$ It follows that $$(R-r)\rho = AB\cdot CE =2 \sqrt{Rr\rho(R-r-\rho)}.$$ Squaring both sides, we get $$(R-r)^2\rho^2 = 4Rr\rho(R-r-\rho),$$ or $$(R-r)^2\rho = 4Rr(R-r) - 4Rr\rho.$$ The desired equality follows from the fact that $$(R-r)^2+4Rr = (R+r)^2.$$

• Thank you so much! You completely solved my question. – jenboo12138 May 1 '17 at 2:29

I assume the following interpretations:

• "Circles touching internally" $\implies$ "Circles intersect" (that is, they overlap),
• "Circle touching the line" $\implies$ "Circle is tangent to the line".

Hints:

• If you haven't yet, draw the circles. The third one will also need to overlap with the previous two for it to touch the line $AB$
• The radius of a circle is always perpendicular to tangent lines
• Join the centers of all circles with straight lines and, together with the radius of the third circle, you'll see why it's a trigonometry question.

Edit: And yes, I'd ask for more clarification for those expressions. Maths usually has quite rigorous definitions to avoid those kind of misinterpretations.