Orthocenter of $\triangle ABC $? Let ABC be a triangle whose vertices are $$A(-5,5)$$ and $$B(7,-1)$$ . If vertex C lies on the circle whose director circle has equation $$x^2 +y^2 = 100$$ then the locus of orthocenter of $$\triangle ABC$$ is ?
I tried solving this by taking a few variables but I am having a hard time eliminating them .

You may notice that $A$ and $B$ are equidistant from the origin, and your hypothesis imply that $C$ lies on a circle centered at the origin through $A$ and $B$. By Euler's theorem, $O,G,H$ are collinear and $OH=3\cdot OG$. Since $A,B$ are fixed and $O$ is the origin,
$$ H = A+B+C $$
lies on a circle centered at $A+B = (2,4)$ having radius $5\sqrt{2}$.

