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)
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.