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?

  • $\begingroup$ By "touching internally", do you mean they intersect (overlap)? $\endgroup$ – AspiringMathematician Apr 26 '17 at 1:27
  • $\begingroup$ 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. $\endgroup$ – jenboo12138 Apr 26 '17 at 1:35
  • $\begingroup$ Are you using the word "touch" to mean "tangent to"? $\endgroup$ – John Wayland Bales Apr 26 '17 at 1:37
  • $\begingroup$ 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. $\endgroup$ – jenboo12138 Apr 26 '17 at 1:45
  • $\begingroup$ 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. $\endgroup$ – 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.

enter image description here

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 $$(R-r)^2+4Rr = (R+r)^2.$$

| cite | improve this answer | |
  • $\begingroup$ Thank you so much! You completely solved my question. $\endgroup$ – 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".


  • 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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.