# Sum of radii of intersecting circles internally tangent to another circle

Let $T$ be a circle with centre at $O$ and radius $R$. Two other circles $T_1$ and $T_2$ with centres at $O_1$ and $O_2$, respectively, are tangent internally to $T$. $T_1$ and $T_2$ intersect one another at the points $A$ and $B$. Find the sum of the radii of $T_1$ and $T_2$, $R_1 + R_2$, if $\angle OAB = \pi/2$. The answer should be expressed in terms of $R$ and the coordinates of the points $O_1$ and $O_2$ (relative to $O$).

(This was a question posted by someone who subsequently deleted his/her account. The question originally generated a lot of comments - and was eventually put on hold - because the OP hadn't made it clear enough. Some of us thought the question is sufficiently interesting to be re-opened so I've cleaned it up and re-posted it anew. The original question is found here but may have been deleted by the time you read this) • Hint: $OO_1BO_2$ is a parallelogram. Aug 13, 2015 at 21:58
• Are you sure that the value of $R_1+R_2$ is fixed? On the face of it, it appears that if the red circle and the blue circle are fixed, it should be possible to construct points $A$ and $B$ in a different position and hence obtain a different value for the radius of the green circle. Aug 14, 2015 at 9:25
• @DavidQuinn I believe so. For fixed blue and red circles and fixed $(A,B)$ points, we only need one more point to uniquely fix the green circle (since it has to pass through both $A$ and $B$). That third point is the point of tangency between the green and blue circles. Aug 14, 2015 at 11:02
• Agreed, but why do points $A$ and $B$ have to be fixed? The condition $OAB=\frac{\pi}{2}$ could be satisfied by many points, which would then determine a different green circle in each case. Aug 14, 2015 at 11:08
• @DavidQuinn Yes, but different choices of $(A,B)$ cause the green circle to have a different centre as well. Hmm... I think I understand now what your concern is. The positions of the red and green circles' centres (relative to the blue circle's centre) must be part of the parameters of the problem. Will amend the question to make that clear. Thanks! Aug 14, 2015 at 11:14

Let me show first of all that, given the two circles $T_1$ and $T_2$ intersecting at $A$ and $B$, there is a unique circle $T$ of center $O$ which is touched internally by both $T_1$ and $T_2$ and such that $\angle OAB=\pi/2$. Let in fact $M$ and $N$ be the tangency points of $T_1$ and $T_2$ respectively with $T$. Then we have $OO_2+O_2N=OO_1+O_1M$, that is: $OO_1-OO_2=R_2-R_1$. As $R_2$ and $R_1$ are given, that means that point $O$ lies on one of the branches of the hyperbola whose foci are $O_1$ and $O_2$ and whose constant difference is $R_2-R_1$ (notice that on the other branch we would have $OO_1-OO_2=R_1-R_2$, so this branch must not be taken into account). In addition, as $\angle OAB=\pi/2$, then $O$ belongs to the line through $A$ parallel to $O_1O_2$. But a branch of hyperbola intersects every line parallel to its major axis at a single point, so the position of point $O$ is always uniquely determined, as stated.

Let's now give a simple construction of this point: $O$ is the point of intersection between the line passing through $O_2$ and parallel to $O_1B$, and the line passing through $A$ and parallel to $O_1O_2$. To prove that, extend $BO_1$ to diameter $BD$ and notice that $D$ belongs to line $OA$. It is then easy to find that triangles $OO_1D$ and $O_1OO_2$ are equal, so that $OO_1BO_2$ is a parallelogram and $OO_2=BO_1=R_1$ and $OO_1=BO_2=R_2$. Extend $OO_2$ to meet $T_2$ at $N$ and $OO_1$ to meet $T_1$ at $M$: we have then $ON=OM$, so we have proven that the circle of center $O$ and radius $R=OM$ touches both $T_1$ and $T_2$.

But from the above construction we immediately get $R_1+R_2=R$, which is the answer to the original question.

• Very nice! +1 for now. I want to see other solutions, if anyone is willing to provide them, before I accept an answer but, wow, very nice. Aug 14, 2015 at 18:02
• The proof is based on the intuition/guess that OO2BO1 is a parallelogram. A more direct way (without needing that intuition) is starting from ON = OM = R, ON = OO2+O2C, OM = OO1 + O1D hence we are looking for O on CD parallel to O1O2 s.t. OO2+O2C = OO1+O1D. Now it is easier to see that O must be the midpoint of CD implying the guessed parallelogram. Aug 15, 2015 at 20:02
• another nice property can be seen from the above diagram: the 3 points N, B, M are on one line. Aug 15, 2015 at 20:04