Circles of Apollonius Let A, B be two distinct points in the plane and $k \in \mathbb{R}_{+}^{*} \backslash\{-1\}$. Determine $\Gamma_{k}=\{M \mid M A=k . M B\}$.
On a:

$$
\begin{aligned}
M & \in \Gamma_{k} \\
\Leftrightarrow \quad \overrightarrow{M A}^{2}-k^{2}(\overrightarrow{M B})^{2} & =0 \\
\Leftrightarrow \quad(\overrightarrow{M A}-k \overrightarrow{M B})(\overrightarrow{M A}+k \overrightarrow{M B}) & =0
\end{aligned}
$$

Let I be the barycenter of (A,1), (B,-k) and J the barycenter of (A, 1), (B,+k). By definition of the barycenter, we have, for any point \( M \) in the plane:

$$
\overrightarrow{M A}-k \overrightarrow{M B}=(1-k) \overrightarrow{M I}, \overrightarrow{M A}+k \overrightarrow{M B}=(1+k) \overrightarrow{M J}
$$

Then:

$$
\begin{aligned}
& M \in \Gamma_{k} \\
& \Leftrightarrow \quad\left(1-k^{2}\right)(\overrightarrow{M I} \cdot \overrightarrow{M J})=0 \\
& \Leftrightarrow \quad \overrightarrow{M I} \cdot \overrightarrow{M J}=0
\end{aligned}
$$

Thus, \(\Gamma_{k}\) is the circle with diameter [IJ].

## 5 Practice Exercises

## 1 Midway Practice

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