Find locus of all points that its sum of squares of distance between them and point $A(x_A,y_A)$ and $B(x_B,y_B)$ equals to $k$.

Question

Find locus of all points that its sum of squares of distance between them and point $A(x_A,y_A)$ and $B(x_B,y_B)$ equals to $k$.

We know that all the points that their sum of distances between them and two points are equal to $k$, form an oval. But this question is about sum of "square of distances", not sum of distances.
Equation of all points in this locus is :
$(x-x_A)^2 + (y-y_A)^2 + (x-x_B)^2 + (y-y_B)^2 = k$
I replaced $x_A,y_A,x_B$ and $y_B$ and $k$ with numbers and drew the graph, and noticed that it is still an oval! Why does this happen? And how to calculate the center of this?
Sorry for bad English.

Answer 1

The equation of all points of this locus is :

$(x-x_A)^2+(y-y_A)^2+(x-x_B)^2+(y-y_B)^2=k\;\;,$

$2x^2\!+\!2y^2\!-\!2(x_A\!+\!x_B)x\!-\!2(y_A\!+\!y_B)y\!+\!x_A^2\!+\!y_A^2\!+\!x_B^2\!+\!y_B^2\!=\!k\;,$

$x^2\!+\!y^2\!-\!(x_A\!+\!x_B)x\!-\!(y_A\!+\!y_B)y\!+\dfrac{x_A^2\!+\!y_A^2\!+\!x_B^2\!+\!y_B^2\!-\!k}2=0\;.$

It is a circle and its center is the point $\,C\left(\dfrac{x_A\!+\!x_B}2,\dfrac{y_A\!+\!y_B}2\right)$ which is the midpoint of the segment $AB$.

Moreover the length of radius is :

$\begin{align}r&=\sqrt{\left(\dfrac{x_A\!+\!x_B}2\right)^2\!+\left(\dfrac{y_A\!+\!y_B}2\right)^2\!-\dfrac{x_A^2\!+\!y_A^2\!+\!x_B^2\!+\!y_B^2\!-\!k}2}=\\[3pt]&=\sqrt{\dfrac{2x_Ax_B+2y_Ay_B-x_A^2-x_B^2-y_A^2-y_B^2+2k}4}=\\[3pt]&=\sqrt{\dfrac{2k-\left(x_A-x_B\right)^2-\left(y_A-y_B\right)^2}4}=\sqrt{\dfrac{2k-\overline{AB}^2}4}=\\[3pt]&=\dfrac12\sqrt{2k-\overline{AB}^2}\;.\end{align}$

Answer 2

It can also be done with Cosine Rule as follows:

In the figure, $M$ is the mid-point of $AB$. Applying cosine rule on $\Delta MPB$,

$$a^2=d^2+\left( \frac{c}{2} \right)^2-2d\left( \frac{c}{2} \right)\cos \alpha \tag{1}$$

Applying cosine rule on $\Delta MPA$

$$b^2 =d^2+\left( \frac{c}{2} \right)^2-2d\left( \frac{c}{2} \right)\cos \left(\pi - \alpha \right) $$

$$b^2 = d^2+\left( \frac{c}{2} \right)^2+2d\left( \frac{c}{2} \right)\cos \left(\alpha \right) \tag{2}$$

$\because a^2+b^2=k$,

$(1)+(2) \implies$ $$a^2+b^2=2d^2+\frac{c^2}{2}$$

$$d=\frac{\sqrt{2k-c^2}}{2}$$

Since $d=\frac{\sqrt{2k-c^2}}{2}$ is a constant, the locus of $P$ is a circle centered at the mid-point $M$ of $AB$ and has radius $d=\frac{\sqrt{2k-c^2}}{2}.$

Answer 3

For sake of matching dimension let $k=c^2$ and consider two foci $(\pm a ,0).$ We have

$$ (x-a)^2+y^2 +(x+a)^2+y^2 =c^2 \to x^2+y^2=c^2/2-a^2 $$

which is the equation of a circle of radius

$$\sqrt{c^2/2-a^2}=\sqrt{k/2-a^2}$$

centered at the origin. For a real circle locus to exist the given $k$ should be

$$ k>\sqrt 2 a. $$