April wrote: In the plane are given a circle $w=C(O,r)$ and a fixed point $A$ outside of this circle. For a mobile point $M\in w$ let $[MN]$ be a diameter of the circle $w$. Denote the circumcircle $w_{m}=C(L)$ of the triangle $AMN$. Ascertain the geometrical locus of the point $L$. Proof. Denote the power $p_{w}(X)$ of the point $X$ w.r.t. the circle $w$. Observe that $O\in MN$- the radical axis of the circles $w_{m}$ and $w$ $\implies$ $p_{w_{m}}(O)=p_{w}(O)$ $\implies$ $LO^{2}-LA^{2}=-r^{2}$ $\implies$ $LA^{2}-LO^{2}=r^{2}$ (constant) In conclusion, the geometrical locus of the point $L$ is a perpendicular line on the fixed line $OA$.

Nice! This line will intersect $OA$ at a point $K$. Let's find $K$ The circle $w_{m}$ intersects the line $OA$ at $A$ and $T$, where $T$ is a fixed point because $OA\cdot OT = r^{2}$ $O$ is between $A,T$. $LA=LT\Rightarrow L$ is on the perpendicular bisector of $AT$ Namely, if $A(a,0)$ then $T\left(-\frac{r^{2}}{a},0\right)$ Then $x_{K}= \frac{a^{2}-r^{2}}{2a}$ Note We have the same solution if we choose the point $A$ in the interior of $w$

this is a very beautiful problem...locus questions are quite difficult to come up with, in my opinion, because there are so few options for the result

Approach by tdl: Call $ I$ is circumcenter of $ (AMN)$, we have: $ AI^2 - OI^2 = MI^2 - OI^2 = OM^2 = R^2$ Then locus of $ I$ is a line perpendicular with $ AO$! Approach by grobber: Let $ P$ be that circumcenter. We have $ R^2 = PM^2 - PO^2 = PA^2 - PO^2$, so $ PA^2 - PO^2$ is constant, meaning that the locus is a line perpendicular to $ AO$ Approach by Pascual2005: Let A' be the inverse of A over an induction with center O and ratio R, and let A'' be the image of A' under the reflection across the center O. Then: A,O,A'' are collinear and M,O,N are collinear, but OA*OA''=OA*OA'=R ² and also OM*ON=R ² wich implies that AMA''N are conciclic and then the circuncircle Is in the perpendicular bisector of AA''. (R=r)

Inversion with center $ O$ and negative power $ - r^2$ (ordinary inversion with positive power $ r^2$ followed by reflection in the inversion center) takes $ A$ into $ A' \in AO,$ such that $ \overline{OA'} \cdot \overline{OA} = - r^2.$ The cirumcircle $ (P)$ of $ \triangle AMN$ cuts the inversion circle $ (O)$ at diametrically opposite points $ \Longrightarrow$ this inversion takes the circle $ (P)$ into itself. Since $ A, A' \in (P),$ locus of circumcenter $ P$ is the perpendicular bisector of $ AA'.$