
Avant tout, faisons une figure.

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\definecolor{zzttqq}{rgb}{0.6,0.2,0}
\definecolor{qqqqff}{rgb}{0,0,1}
\definecolor{xdxdff}{rgb}{0.49,0.49,1}
\definecolor{cqcqcq}{rgb}{0.75,0.75,0.75}

\begin{center}
\begin{tikzpicture}[>=latex,scale=1.1]
\draw [color=cqcqcq,dash pattern=on 2pt off 2pt, xstep=1.0cm,ystep=1.0cm] (-3.24,-1.33) grid (10.58,10.45);
\draw[->,color=black] (-3.24,0) -- (10.58,0);
\foreach \x in {-3,-2,-1,1,2,3,4,5,6,7,8,9,10}
\draw[shift={(\x,0)},color=black] (0pt,2pt) -- (0pt,-2pt) node[below] {\footnotesize $\x$};
\draw[->,color=black] (0,-1.33) -- (0,10.45);
\foreach \y in {-1,1,2,3,4,5,6,7,8,9,10}
\draw[shift={(0,\y)},color=black] (2pt,0pt) -- (-2pt,0pt) node[left] {\footnotesize $\y$};
\draw[color=black] (0pt,-10pt) node[right] {\footnotesize $0$};
\clip(-3.24,-1.33) rectangle (10.58,10.45);
\fill[color=zzttqq,fill=zzttqq,fill opacity=0.1] (0,4) -- (3,6) -- (6,0) -- cycle;
\fill[draw=qqqqff,fill=qqqqff,fill opacity=0.1] (0.44,4.3) -- (1.15,3.23) -- (3.74,4.52) -- cycle;
\draw [color=zzttqq] (0,4)-- (3,6);
\draw [color=zzttqq] (3,6)-- (6,0);
\draw [color=zzttqq] (6,0)-- (0,4);
\draw [domain=-3.24:10.58] plot(\x,{(-6-6*\x)/-4});
\draw [domain=-3.24:10.58] plot(\x,{(--24-4*\x)/6});
\draw(2.08,4.62) circle (1.66cm);
\fill [color=xdxdff] (0,4) circle (1.5pt);
\draw[color=xdxdff] (0.12,4.2) node {$A$};
\fill [color=qqqqff] (3,6) circle (1.5pt);
\draw[color=qqqqff] (3.26,6.13) node {$B$};
\fill [color=xdxdff] (6,0) circle (1.5pt);
\draw[color=xdxdff] (6.13,0.2) node {$C$};
\fill [color=uququq] (1.15,3.23) circle (1.5pt);
\draw[color=uququq] (1.23,3) node {$H$};
\fill [color=uququq] (0.44,4.3) circle (1.5pt);
\draw[color=uququq] (0.57,4.5) node {$E$};
\fill [color=uququq] (3.74,4.52) circle (1.5pt);
\draw[color=uququq] (3.85,4.72) node {$F$};
\end{tikzpicture}
\end{center}
\begin{enumerate}
\item $\begin{aligned}[t]
\prodscal{AC}{BH}=0 & \Longleftrightarrow \dbinom{6}{-4}\cdot\dbinom{x_H-3}{y_H-6}=0\\
& \Longleftrightarrow 6(x_H-3)-4(y_H-6)=0
\end{aligned}$

$\begin{aligned}[t]
\prodscal{AC}{BH}=0 & \Longleftrightarrow 6x_H-4y_H+6=0\\
& \Longleftrightarrow 3x_H-2y_H+3=0\qquad (1)
\end{aligned}$

De plus, on sait que $\vv{u}\dbinom{x}{y}$ et $\vv{v}\dbinom{x'}{y'}$ sont colinéaires si et seulement si $xy'-x'y=0$, donc :
\begin{align*}
\vv{\text{AH}}\text{ et }\vv{\text{AC}}\text{ sont colinéaires}
& \Longleftrightarrow -4x_H-6(y_H-4)=0\\
& \Longleftrightarrow -4x_H-6y_H+24=0\\
& \Longleftrightarrow 2x_H+3y_H-12=0\qquad(2)
\end{align*}
On a alors : $\begin{aligned}[t]\begin{cases}
3x_H-2y_H+3=0 & (1)\\
2x_H+3y_H-12=0 &(2)
\end{cases}
&
\Longleftrightarrow
\begin{cases}
6x_H-4y_H+6=0 & 2\times(1)\rightarrow L_1\\
6x_H+9y_H-36=0 & 3\times(1)\rightarrow L_2
\end{cases}\\
& \Longrightarrow 13y_H-42=0\qquad L_2-L_1\\
& \Longrightarrow  y_H=\dfrac{42}{13}\\
& \Longrightarrow 2x_H=12-3\times\dfrac{42}{13}\qquad\text{d'après (2)}\\
& \Longrightarrow x_H=6-\dfrac{63}{13}=\dfrac{15}{13}
\end{aligned}$

\underline{Ainsi, H\coord{\dfrac{15}{13}}{\dfrac{42}{13}}.}

\item $\begin{aligned}[t]
\text{M}\coord{x}{y}\in\mathcal{C} & \Longleftrightarrow \prodscal{MB}{MH}=0\\
& \Longleftrightarrow \dbinom{3-x}{6-y}\cdot\dbinom{\frac{15}{13}-x}{\frac{42}{13}-y}=0\\
& \Longleftrightarrow (3-x)\left(\dfrac{15}{13}-x\right)+(6-y)\left(\dfrac{42}{13}-y\right)=0\\
& \Longleftrightarrow x^2+y^2-\dfrac{54}{13}x-\dfrac{120}{13}y+\dfrac{297}{13}=0
\end{aligned}$

\underline{Une équation cartésienne de $\mathcal{C}$ est donc $x^2+y^2-\dfrac{54}{13}x-\dfrac{120}{13}y+\dfrac{297}{13}=0$.}

\item 
\begin{itemize}
\item Déterminons d'abord les coordonnées \coord{x}{y} du point E, intersection de $\mathcal{C}$ et [AB[.

$\begin{aligned}[t]
\text{E}\coord{x}{y}\in[\text{AB}[ & \Longleftrightarrow \vv{\text{AE}}\text{ et }\vv{\text{AB}}\text{ sont colinéaires}\\
& \Longleftrightarrow \begin{vmatrix}
x & 3	\\
y-4 & 2
\end{vmatrix}=0\\
& \Longleftrightarrow 2x-3y+12=0\\
& \Longleftrightarrow x=\dfrac{3}{2}y-6
\end{aligned}$

De plus, $\text{E}\in\mathcal{C}$ donc :
\begin{align*}
 & x^2+y^2-\dfrac{54}{13}x-\dfrac{120}{13}y+\dfrac{297}{13}=0\\
\Longleftrightarrow & \left(\dfrac{3}{2}y-6\right)^2+y^2-\dfrac{54}{13}\left(\dfrac{3}{2}y-6\right)-\dfrac{120}{13}y+\dfrac{297}{13}=0
\end{align*}

\begin{align*}
\Longleftrightarrow &  169y^2-\numprint{1740}y+\numprint{4356}=0\\
\Longleftrightarrow & y=\dfrac{726}{169}\qquad\text{ou}\qquad y=6
\end{align*}

Or, $\text{E}\neq\text{B}$ donc $y=\dfrac{726}{169}$ et donc $x=\dfrac{3}{2}\times\dfrac{726}{169}-6=\dfrac{75}{169}$.

\underline{Ainsi, E\coord{\dfrac{75}{169}}{\dfrac{726}{169}}.}


\item Déterminons ensuite les coordonnées \coord{x}{y} du point F, intersection de $\mathcal{C}$ et ]BC].

$\begin{aligned}[t]
\text{F}\coord{x}{y}\in]\text{BC}] & \Longleftrightarrow \vv{\text{BF}}\text{ et }\vv{\text{BC}}\text{ sont colinéaires}\\
& \Longleftrightarrow \begin{vmatrix}
x-3 & 3	\\
y-6 & -6
\end{vmatrix}=0\\
& \Longleftrightarrow -6(x-3)-3(y-6)=0\\
& \Longleftrightarrow -6x-3y+36=0\\
& \Longleftrightarrow y=12-2x
\end{aligned}$

\medskip

De plus, $\text{F}\in\mathcal{C}$ donc :
\begin{align*}
 & x^2+y^2-\dfrac{54}{13}x-\dfrac{120}{13}y+\dfrac{297}{13}=0\\
 \Longleftrightarrow & x^2+(12-2x)^2-\dfrac{54}{13}x-\dfrac{120}{13}(12-2x)+\dfrac{297}{13}=0\\
 \Longleftrightarrow & 65x^2-438x+729=0\\
 \Longleftrightarrow & x=3\qquad\text{ou}\qquad x=\dfrac{243}{65}
\end{align*}

Or, $\text{F}\neq\text{B}$ donc $x=\dfrac{243}{65}$, et donc $y=12-2\times\dfrac{243}{65}=\dfrac{294}{65}$.

\underline{Ainsi, F\coord{\dfrac{243}{65}}{\dfrac{294}{65}}.}

\item Regardons alors si (HE) et (HF) sont perpendiculaires.

\begin{align*}
\prodscal{HE}{HF} & = \dbinom{\frac{75}{169}-\frac{15}{13}}{\frac{726}{169}-\frac{42}{13}}\cdot\dbinom{\frac{243}{65}-\frac{15}{13}}{\frac{294}{65}-\frac{42}{13}}\\
& = \dbinom{-\frac{120}{169}}{\frac{180}{169}}\cdot\dbinom{\frac{168}{65}}{\frac{84}{65}}\\
& = -\frac{120}{169}\times\dfrac{168}{65}+\dfrac{180}{169}\times\dfrac{84}{65}\\
& = -\dfrac{\numprint{1008}}{\numprint{2197}}\\
\prodscal{HE}{HF} & \neq 0.
\end{align*}
\underline{Par conséquent, le triangle EHF n'est pas rectangle en F.}

\item $\begin{aligned}[t]
\prodscal{HE}{HF} & = \|\vv{\text{HE}}\|\times\|\vv{\text{HF}}\|\times\cos\angleor{HE}{HF}\\
-\dfrac{\numprint{1008}}{\numprint{2197}} & = \sqrt{\left(-\dfrac{120}{169}\right)^2+\left(\dfrac{180}{169}\right)^2}\times\sqrt{\left(\dfrac{168}{65}\right)^2+\left(\dfrac{84}{65}\right)^2}\times\cos\angleor{HE}{HF}\\
\cos\angleor{HE}{HF} & = \dfrac{-\dfrac{\numprint{1008}}{\numprint{2197}}}{\sqrt{\left(-\dfrac{120}{169}\right)^2+\left(\dfrac{180}{169}\right)^2}\times\sqrt{\left(\dfrac{168}{65}\right)^2+\left(\dfrac{84}{65}\right)^2}}\\
\cos\angleor{HE}{HF} &  \approx \numprint{0,124}\\
\Aboxed{\angleor{HE}{HF} & \approx 97\degres}
\end{aligned}$
\end{itemize}
\end{enumerate}
