It's somehow intresting to me that our teacher gave us this problem about $2$ years ago! for solving it, extend $PC$ to intersect the circle with center $O_1$...

Hmm, this is basically question 2 of 15 IBMO-Venezuela. See the following threads http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=83793 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=336494 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=288&t=80971

This solution assumes $AB$ is the closer tangent to $P$, which I think the original problem states (it doesn't matter though). Let $(O_1)\cap(O_2)=\{P,Q\}$, $M=AB\cap PQ$, and $A'$ be the reflection of $A$ over $O_1$. Clearly $\angle{CPA}=90^\circ\Longleftrightarrow\angle{AA'C}=\angle{AA'P}$. But $\angle{AA'P}=\angle{BAP}$, so it suffices to show $\triangle{AA'C}\sim\triangle{BAP}\Longleftrightarrow\triangle{AO_1C}\sim\triangle{BMP}$ (equivalent since $O_1,M$ are the midpoints of $AA',BA$ respectively), which is just a simple angle chase.

RaleD wrote:

This solution by RaleD is a very nice solution. At first glance, I thought it is similar to mine. In fact, it's way simpler and nicer. In my opinion, it's very clever to use the radical axis. My solution also uses something like the equation $CN*CA=CP^2$ to get a geometric relation. Here is how it goes: Let the line through $C$ perpendicular to $AB$ intersect $AB$ at point $S$. A direct calculation of side lengths (starting with $BC^2-PC^2=AB^2-AP^2$) yields $AS\cdot AB=AP^2$ Now, it is easy to find that $A,S,C,P$ are concyclic, which leads to $\angle{APC}=90^\circ$

Denote $Q$ the other intersection point of two circles,let $M$ be the intersection point of $PQ$ and $AB$ and let $O$ be the circumcenter of $ABQ$.Now,let $C'$ be the intersection of the perpendicular from $A$ to $BP$ and the perpendicular from $P$ to $AP$.Now,we have that $AOM$ is similar to $APC'$ so we get $APM$ is similar to $AC'P$.Now,from an easy angle chase and using the previos similarity we get $QC'O$ is similar to $QPB$ and from this we get $QC'=PC'$ so $C'=C$ and that is it.

Here's another solution: Define $Z \equiv AB \cap O_1O_2$ and let $ZP, AC$ cut $(O_1)$ for a second time at $Q, C'$, respectively. Let $H$ be the projection of $A$ onto $O_1O_2.$ Note that $Z$ is the external center of homothety that maps $(O_2) \mapsto (O_1).$ It is clear that this homothety takes $B \mapsto A$ and $P \mapsto Q$, so $AC \perp BP \implies AC \perp AQ.$ Hence, $\overline{QC'}$ is a diameter of $(O_1) \implies Q, C', O_1$ are collinear. Furthermore, note that $\angle ZHA = \angle ZAO_1 = 90^{\circ} \implies \triangle ZAH \sim \triangle ZO_1A \implies ZH \cdot ZO_1 = ZA^2 = ZP \cdot ZQ$, where the last step follows from Power of a Point. Therefore, $H, O_1, P, Q$ are concyclic, so $\measuredangle PHC = \measuredangle PHO_1 = \measuredangle PQO_1 = \measuredangle PQC' = \measuredangle PAC' = \measuredangle PAC$, where the angles are directed. Hence, $P, H, C, A$ are concyclic, so $\measuredangle APC = \measuredangle AHC = 90^{\circ}$, as desired. $\square$

with using the special point on the median ( X ) , the problem sounds easy

My solution is pretty much the identical solution outlined through by @math154. Let $P, Q$ be the points of intersection between circles with circumcenters $O_1, O_2$ with $P$ closer to line $AB$ than $Q$. Let $X = PQ \cap O_1O_2$, and $A_1, B_1$ denote the reflections of $A, B$ over $O_1, O_2$, respectively. Also let $M$ denote the midpoint of line $AB$. We then make the following claim: Claim: $\triangle A_1O_1C \sim \triangle AMP$. Proof: We have that as $AA_1 \parallel BB_1$, that \[\measuredangle A_1OC = \measuredangle BO_2C = \measuredangle AMP, \]as $BMXO_2$ is also cyclic. $\square$ We now show that $A_1, C, P$ are collinear: Claim: $A_1, C, P$ are collinear. Proof: We wish to show $\measuredangle AA_1C = \measuredangle AA_1P$. It then suffices to equivalently show $\triangle A_1O_1C \sim \triangle AMP$. Indeed, we have that \[\measuredangle AA_1C = \measuredangle MAP, \]and by the above claim/lemma, we obtain the desired similarlity. $\square$ Hence we have that $A_1, C, P$ are collinear, in particular $AP \perp PC$, as desired. $\blacksquare$

Let $T$ the symmetric of $A$ wrt $O_1O_2$ ; $S$ be the ex-similicenter ; $Q$ the intersection of $(O_1)$ and $ SP$; $P',A'$ the antipodes of $P,A$; $D$ the intersection of $A'P'$ and $O_1O_2$ ; $C'$ the intersection of $ PP'$ and $O_1O_2$ we will show that $C'=C$ : by butterfly we deduce that $O_1C'=O_1D$ i.e. $PAP'A'$ is parallelogram. More $A'(PQ,AT)=-1$ but $A'(C'D,O_1 T)=-1$ hence $A',D,Q$ are collinear. Since $A'Q\parallel AC'$ then $AC'\perp AQ\parallel BP$ therefore $C'=C$ .

WakeUp wrote: Let one of the intersection points of two circles with centres $O_1,O_2$ be $P$. A common tangent touches the circles at $A,B$ respectively. Let the perpendicular from $A$ to the line $BP$ meet $O_1O_2$ at $C$. Prove that $AP\perp PC$.

which is quite neat: In triangle $ABC$ with orthocentre $H$ and point $Q$ on arc $\widehat{BAC}$ such that $\angle AQH =90^{\circ}$ (hence $QH$ bisects $BC$), the line through $H$ parallel to $AB$, the line $AC$, and the perpendicular bisector of line $QH$ concur. Let $X, Y$ be points on $AB, AC$ such that $AXHY$ is a parallelogram. Then line $XY$ passes through the midpoint of $AH$, i.e, the centre of the circumcircle of $\triangle AQH$. Also, $(AX, AY, AH, AQ)=-1$ by projecting all on line $BC$, hence $XY \parallel AQ$ or $XY \perp QH$, hence $XY$ is the perpendicular bisector of $QH$, and proving $Y$ lies on it was all we needed.