Locus using Euclidean geometry

CONSTRUCTION :

Given the triangle ABC: M is the middle of AB. C' is symmetric of C relatively to M.

Draw the semi-circle $(C_1)$ of diameter CC'.

Draw an arc of circle $(C_2)$ of center C and radius=$\overline{AB}$.

$(C_1)$ and $(C_2)$ intersect in U.

Draw the semi-circle $(C_3)$ of diameter C'U.

V is the middle of C'U. From V, draw a straight line orthogonal to C'U. It intersects $(C_3)$ in W.

Draw the circle $\color{red}{(C)}$ of center C' and radius=$\overline{C'W}$

From right triangle CUC' :$\quad\overline{C'U}^2=\overline{C'C}^2-\overline{AB}^2$

From right triangle C'WU :$\quad 2\overline{C'W}^2=\overline{C'U}^2$

$$\overline{C'W}^2=\frac{1}{2}\left(\overline{C'C}^2-\overline{AB}^2 \right)$$

$$\overline{C'W}^2=2\left(\overline{C'M}^2-\overline{MB}^2 \right) \tag 1$$

We will show that the circle $\color{red}{(C)}$ of center C' and radius=$\overline{C'W}$ is the locus of the points P such that $$\overline{PA}^2+\overline{PB}^2=\overline{PC}^2$$

PROOF :

H is the orthogonal projection of C on AB. H' is symmetric of H relatively to M.

First, consider an arbitrary point P , not necessarily on $\color{red}{(C)}$

J, K, G are the orthogonal projections of P respectively on AB, CH, C'H' (or on straight extension).

The Pythagorean theorem in the respective right triangles gives :

$\overline{PA}^2=(\overline{AM}+\overline{MJ})^2+\overline{JP}^2$

$\overline{PB}^2=(\overline{MB}-\overline{JB})^2+\overline{JP}^2$

$\overline{PC}^2=\overline{PK}^2+\overline{KC}^2=(\overline{MH}-\overline{MJ})^2+(\overline{HC}-\overline{JP})^2 $

Second, consider a particular point P with condition : $$\overline{PA}^2+\overline{PB}^2=\overline{PC}^2$$ $(\overline{AM}+\overline{MJ})^2+\overline{JP}^2+(\overline{MB}-\overline{MJ})^2+\overline{JP}^2=(\overline{MH}-\overline{MJ})^2+(\overline{HC}-\overline{JP})^2$

After expending and few transformations :

$(\overline{MH}+\overline{MJ})^2+(\overline{HC}+\overline{JP} )^2=2\overline{MH}^2+2\overline{HC}^2-\overline{AM}^2-\overline{MB}^2$

$$(\overline{MH}+\overline{MJ})^2+(\overline{HC}+\overline{JP} )^2=2\overline{MH}^2+2\overline{HC}^2-2\overline{MB}^2$$

$\overline{MH}+\overline{MJ}=\overline{H'M}+\overline{MJ}=\overline{H'J}=\overline{GP}$ and $\overline{HC}+\overline{JP}=\overline{C'H'}+\overline{H'G}=\overline{C'G}$

$$\overline{GP}^2+\overline{C'G}^2=2\overline{MH}^2+2\overline{HC}^2-2\overline{AB}^2$$ In the right triangle C'GP : $\overline{GP}^2+\overline{C'G}^2= \overline{C'P}^2$

$$\overline{C'P}^2=2\overline{MH}^2+2\overline{HC}^2-2\overline{MB}^2$$ $$\overline{C'P}^2= 2\overline{MC}^2-2\overline{MB}^2 = 2\overline{C'M}^2-2\overline{MB}^2$$

A, B, C, H, M are fixed points. Hence $\quad 2\overline{C'M}^2-2\overline{MB}^2$ = Constant.

$\overline{C'P}$ is of constant lengh. Thus the point P is on a circle of center C' and radius $R$. This is the locus. $$R^2=2\overline{C'M}^2-2\overline{MB}^2$$

Compare this result to the previous (1) above : $\quad\overline{C'W}^2=2\left(\overline{C'M}^2-\overline{MB}^2 \right)$

Indeed, circle $\color{red}{(C)}$ defined on the first figure is the locus.

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 $r$.