[ Method of coordinates on the plane ]

Given points $A, B$ and a positive number $d$. Find the geometric locus of points $M$, for which $A M^{2}+B M^{2}=d$.

#
Choose the coordinate system $X O Y$ so that point $A$ is its origin, and point $B$ lies on the positive half-axis $O X$.

## Solution

Let $A B=b$. Choose the coordinate system $X O Y$ so that point $A$ is its origin, and point $B$ lies on the positive half-axis $O X$. Then the coordinates of points $A$ and $B-(0 ; 0)$ and $(b ; 0)$. Point $M(x ; y)$ belongs to the desired geometric locus if and only if

$$
\begin{gathered}
A M^{2}+B M^{2}=d \Leftrightarrow x^{2}+y^{2}+(x-b)^{2}+y^{2}=d \Leftrightarrow \\
\Leftrightarrow 2 x^{2}-2 b x+2 y^{2}+b^{2}=d \Leftrightarrow x^{2}-b x+y^{2}=\frac{d-b^{2}}{2} \Leftrightarrow \\
\Leftrightarrow\left(x-\frac{b}{2}\right)^{2}+y^{2}=\frac{d-b^{2}}{2}+\frac{b^{2}}{4} \Leftrightarrow\left(x-\frac{b}{2}\right)^{2}+y^{2}=\frac{2 d-b^{2}}{4}
\end{gathered}
$$

If $d>\frac{b^{2}}{2}$, then the desired geometric locus is a circle with radius $\frac{1}{2} \sqrt{2 d-b^{2}}$ centered at the midpoint of segment $A B$. If $d=\frac{b^{2}}{2}$, then it results in a single point - the midpoint of segment $A B$. If $d<\frac{b^{2}}{2}$, then the desired geometric locus is empty.

| Classes: 7,8,9 |
| :---: | :---: | :---: |
|  | Integer lattices (other). |  |
|  | Colorings |  |
|  | Geometry on grid paper |  |

If $d>\frac{A B^{2}}{2}$, then the desired GML is a circle centered at the midpoint of segment $A B$.