TITLE: Locus using Euclidean geometry
QUESTION [10 upvotes]: 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?

REPLY [1 votes]: Let's take $C$ as origin of a point vectors, so $\overrightarrow{C}=\overrightarrow{0}$. Then we have: 
$$\overrightarrow{PA}^2 + \overrightarrow{PB}^2 =
\overrightarrow{PC}^2$$ 
$$(\overrightarrow{A}-\overrightarrow{P})^2 + (\overrightarrow{B}-\overrightarrow{P})^2 =
\overrightarrow{P}^2$$
$$\overrightarrow{P}^2-2(\overrightarrow{A} + \overrightarrow{B})\overrightarrow{P}+\overrightarrow{A}^2+\overrightarrow{B}^2
=\overrightarrow{0} $$
Let $\overrightarrow{Q} =
\overrightarrow{A}+\overrightarrow{B}$, so $Q$ is such a point that $AQBC$ is a parallelogram. Now we have
$$\overrightarrow{P}^2-2\overrightarrow{Q}\cdot  \overrightarrow{P} +\overrightarrow{Q}^2-2\overrightarrow{A}\cdot \overrightarrow{B}
=\overrightarrow{0} $$
$$(\overrightarrow{P}-\overrightarrow{Q})^2=2\overrightarrow{A}\cdot \overrightarrow{B} $$
Now suppose that $\angle ACB<90^{\circ}$ and let $r= \sqrt{2\overrightarrow{A}\cdot \overrightarrow{B}}$. Then we have:
$$ QP^2=r^2$$
which means that $Q$ describes a circle with a center at $Q$ and a radius $r$.
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