Find the radius of the third circle given three circles

Question

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?

Answer 1

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

Answer 2

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".

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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.

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