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

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.

• You seem to assume circumradius $=1$, is that so? – Aretino Nov 27 '18 at 15:32
• @Aretino, yes , just for the sake of simplicity. – Awe Kumar Jha Nov 27 '18 at 15:33

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).