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)

 A: 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.
