Let $P$ be any point in the plane. Find the locus of $P$ such that $PA^2 + PB^2 = PC^2$, where $ABC$ is a triangle.
I have found the locus. It's a circle having center at point $Q$, such that $AQBC$ is a parallelogram. Now from $A$, drop a perpendicular $AD$ on $BC$. Taking $BC$ as diameter, draw a circle and say it intersects $AD$ at $K$. Draw $MC$ perpendicular to $KC$ and equal to $KC$. Take the length $KM$ and draw a circle having radius equal to $KM$ with its center at at $Q$. This circle represents the locus of $P$. I have found the proof using coordinate geometry. Can someone suggest a solution using pure Euclidean geometry?