Prove that locus of vertex is $(a+b)(x^2+y^2)+2h(x\beta + \alpha y) + (a-b)(x\alpha - y\beta)=0$

Question

The base of a triangle passes through a fixed point $(\alpha ,\beta )$. Let the perpendicular bisectors of the sides be the lines $ax^2+2hxy+by^2=0$. It is to prove that the locus of the vertex is : $$(a+b)(x^2+y^2)+2h(x\beta + \alpha y) + (a-b)(x\alpha - y\beta)=0$$ Clearly, the origin is the circumcentre of the triangle. So , it is easy to take polar coordinates and define, $$x:=\cos \theta , y:=\sin \theta$$ $$\alpha :=\cos \phi , \beta :=\sin \phi$$ $$\tan \psi = \frac {a-b}{2h}$$ This greatly simplifies the desired expression to, $$(a+b)+2h\sec \psi \sin {(\theta + \phi + \psi)}=0$$ Yet this simplification is useless, unless I have a relation between the point through the base and the vertex. Any suggestions are welcome.

Answer 1

I'll describe here a construction of the triangle: you could possibly use it to obtain the equation of the locus.

Let $AB$ be the base of the triangle (containing point $P=(\alpha,\beta)$) and $C$ its third vertex. Notice that the angle $\psi$ between the perpendicular bisectors of $AC$ and $BC$ (red and blue dashed lines in the diagram) is the same as $\angle ACB$ and also the same as $\angle AOQ$, where $OQ$ is the perpendicular bisector of base $AB$ and $O$ the circumcenter of triangle $ABC$.

To construct the triangle, choose then line $OQ$ at will and draw $OQ'$ such that $\angle QOQ'=\psi$. Drop from $P$ the perpendicular to line $OQ$, which will meet line $OQ'$ at $A$. Reflect then $A$ about $OQ$ to get $B$, and reflect either $A$ or $B$ about one of the perpendicular bisectors to get $C$.

As line $OQ$ varies, point $C$ varies too and its locus is indeed a circle (purple circle below).