(a) let $ D: AO_1 \cap C_1$ and $ C: BO_2 \cap C_2$. Cause T is the internal center of similitude of $ C_1$ and $ C_2$ we have $ BC \parallel DA$, so AB pass trought the external center of similitude of $ C_1$ and $ C_2$ that we call Z. (b) call M the midpoint of AB. With an homothety with center Z and ratio $ \frac{r_1+r_2}{2r_1}$, A go in M, and so the locus of point M when A moves on $ C_1$ is a circle with diameter $ O_1O_2$.

My opinion: a) Let $C_1 \left( {K,R} \right)\;and\;C_2 \left( {L,r} \right)$ the circles. Let TM the public tangent of the circles, ${\rm M} \in {\rm A}{\rm B} \Rightarrow KA\parallel LB\;\left( {\mathop {KTA}\limits^ \wedge = \mathop {MTB}\limits^ \wedge = \mathop {TEB}\limits^ \wedge ,and\,\vartriangle KTA,\vartriangle LBE\;isosceles\;triangles} \right).$ The point $S \equiv KL \cap AB$ is constant because is the sender of omothesia of the circles $C_1 \left( {K,R} \right)\;and\;C_2 \left( {L,r} \right).$ $H \equiv AB \cap C_2 ,TE = 2r.$ b) Te KABL is trapezium. Let Q the mint point of the KL, R the mint point of the KL, when $QR = \frac{{R + r}}{2}...$ S.E.Louridas

nice and easy Problem.

Another solution: Let the common external tangent of the two circles be $CD, CD$ cuts $O_{1}O_{2}$ at $K; O_{1}O_{2}$ cuts the two circles at another points $G$ and $E$. $AB$ cuts the two circles at another points $N$ and $M$. Since $\angle CO_{1}G = DO_{2}T$ and both $\Delta CO_{1}G$ and $DO_{2}T$ are isoscales, $\angle KCG = 1/2.\angle CO_{1}G = 1/2.\angle DO_{2}T = \angle DEG$, hence $CGED$ is cyclic. As $AT\perp TB, AG\parallel TB$, thus easy to see that $AO_{1}\parallel BO_{2}$ and $AGEM$ is cyclic. $\angle KCA = 1/2.\angle CO_{1}A = 1/2.\angle DO_{2}B = \angle DMA$, thus $CAMD$ is cyclic. As $CGED$ is cyclic, $KC. KD = KG. KE$, hence $K$ is on the radical axis of the circumcircle of $CAMD$ and the circumcircle of $AGEM$, so $K$ must be on their common chord $AM$. Answer to the first question: All such lines $AB$ pass through $K$. Let $P$ be the midpoint of $AB$. Since $AGTB$ is a trapezium, $PO_{1}\parallel AG$, similarly, $PO_{2}\parallel AT$, hence $\angle O_{1}PO_{2} = 90^{\circ}$. Answer to the second question: The midpoint $P$ of $AB$ lies on the circle with diameter $O_{1}O_{2}$.

Another way is with the all-migthy coordinates. Part a. Let $O_1$ and $O_2$ be the centers of the circles, then we can say that w.l.o.g. $T=(0,0)$, and $O_1=(-y,0)$ and $O_2=(x,0)$, where $y$ is a positive real number. Then we have that: $$c_1:(a+y)^2+b^2=y^2$$$$c_2:(a-x)^2+b^2=x^2$$ Now let $A=(m,n) \in c_1$ and let $l_1,l_2$ be the lines $AT$ and $BT$, respectively. We easily have that: $$l_1:y=\frac{n}{m}x$$$$l_2:y=-\frac{m}{n}x$$this we have since $A \in l_1$ and $l_1 \perp l_2$. Now let $B=(p,q) \in c_2$, since we have that $B \in l_2$ we must have that $q=-\frac{m}{n}p$, and we have that: $$(p-x)^2+q^2=x^2$$this easily implies that: $$\left(1+\frac{m^2}{n^2}\right)p^2-2px=0$$since $p \neq 0 $, we get that: $$p=\frac{2xn^2}{m^2+n^2}$$ Now we need to get the line $AB$. We easily get by calculation that: $$\overline{AB}=l_3:y=-\frac{2xmn+m^2n+n^3}{2xn^2-m^3-mn^2}x+\frac{2xn(n^2+m^2)}{2xn^2-m^3-mn^2}$$now let's look at the intersection with the $x$ axis. We have that the point of intersection has the coordinates: $$x'=\frac{2x(n^2+m^2)}{2xm+m^2+n^2}=-2\frac{xy}{x-y}$$we used that $m^2+n^2=-2my$. Thus we see that the point of intersection with the $x$ axis has now dependency on the points $A$ and $B$, only on th radii of the circles. Thus we have that the lines of type $AB$ intersection all in one point. Part b. We have that $M$ is the circumcenter of $ATB$. and we easily have that $O_1M$ is the perpendicular bisector of $AT$ and $O_2M$ is the perpendicular bisector of $BT$. Thus we have that $\angle O_1MO_2 = 90$, thus we have that $M$ moves along a circle with diameter of $O_1O_2$.

brianchung11 wrote: Two circles $ C_1,C_2$ with different radii are given in the plane, they touch each other externally at $ T$. Consider any points $ A\in C_1$ and $ B\in C_2$, both different from $ T$, such that $ \angle ATB = 90^{\circ}$. (a) Show that all such lines $ AB$ are concurrent. (b) Find the locus of midpoints of all such segments $ AB$. Let $O_1 $ and $O_2$ be the centers of $\mathcal{C}_1$ and $\mathcal{C}_2$, respectively. Denote by $A'$ and $B'$ the antipodes of $A$ and $B$ in $\mathcal{C}_1$ and $\mathcal{C}_2$, respectively. Let $M$ be the midpoint of segment $AB$, and let $N$ be the midpoint of segment $A'B'$. Also, let $X=AB\cap A'B'$. Due to the angle condition, it holds $A-T-B'$ and $B-T-A'$ collinear. By simple angle chasing using the common tangent, it holds $\measuredangle BB'T=\measuredangle A'AT$, so $AA'\parallel BB'$. Also, $O_1-T-O_2-X$ are concurrent, because $$ -1=(A,A';BB'\cap AA',XT\cap AA')=(A,A';\infty_{AA'},XT\cap AA') \Longrightarrow O_1-T-X \thickspace \text{collinear.} $$a) All lines $AB$ concur in $AB\cap A'B'$, which is a fixed point, because $$ -1=(A,A';\infty_{AA'},O_1)\stackrel{B}=(X,T;O_2,O_1) $$b) The locus is the circle with diameter $\overline{O_1O_2}$, except for $O_1$ and $O_2$, because $$ \measuredangle ATB=90^o \Longrightarrow AT\perp BT \Longrightarrow AB'\perp BA' \Longrightarrow MO_2\perp MO_1. $$

brianchung11 wrote: Two circles $ C_1,C_2$ with different radii are given in the plane, they touch each other externally at $ T$. Consider any points $ A\in C_1$ and $ B\in C_2$, both different from $ T$, such that $ \angle ATB = 90^{\circ}$. (a) Show that all such lines $ AB$ are concurrent. (b) Find the locus of midpoints of all such segments $ AB$. Took less than 7 mins but its so fun when u realice what is the problem asking for. (a) Let $O_1,O_2$ the centers of $C_1,C_2$ respectivily, then clearly $O_1,T,O_2$ are colinear and by angle chase u have $\angle TO_1A+\angle TO_2B=180$ which means that $O_1A \parallel O_2B$, now let $O_1T \cap C_1=C$ and $O_2T \cap C_2=D$ then $\angle CAT=\angle ATB=\angle TBD=90$ so $CA \parallel TB$ and $BD \parallel TA$ meaning that $\triangle CAT \sim \triangle DBT$ and since $C,D,T$ are fixed we have that all lines $AB$ pass through the center of those similarities, which is the exsimilicenter of $C_1,C_2$. (b) Since $M$ is the center of $(TAB)$ we have that $O_1M \perp AT$ and $O_2M \perp BT$ so $\angle O_1MO_2=90$ becuase of the rectangle formed by the lines $O_1M,O_2M,AT,BT$ and this tells that the locus of all the points $M$ is the circle with diameter $O_1O_2$ thus we are done