By differentiating implicitly with respect to $x$ both sides of the implicit equation
$$
\begin{equation*}
2x^{2}+y^{2}=33,\tag{1}
\end{equation*}
$$
since the derivatives of both sides should be equal we get successively:
$$
\begin{eqnarray*}
&&\frac{d}{dx}\left( 2x^{2}+y^{2}\right) =\frac{d}{dx}\left( 33\right) \\
&\Rightarrow &\frac{d}{dx}\left( 2x^{2}+y^{2}\right) =0 \\
&\Leftrightarrow &\frac{d}{dx}\left( 2x^{2}\right) +\frac{d}{dx}\left(
y^{2}\right) =0 \\
&\Leftrightarrow &4x+2y\frac{dy}{dx}=0,\qquad \frac{d}{dx}\left(
y^{2}\right) =2y\frac{dy}{dx}\text{ by the chain rule} \\
&\Leftrightarrow &\frac{dy}{dx}=-\frac{4x}{2y}=-\frac{2x}{y}\tag{2} \\
&\Rightarrow &\left. \frac{dy}{dx}\right\vert _{x=2,y=5}=-\frac{4}{5}.\tag{3}
\end{eqnarray*}
$$
The equation of the tangent line at $(2,5)$ is
$$
\begin{equation*}
y-5=-\frac{4}{5}(x-2),\tag{4}
\end{equation*}
$$
while the equation of the normal line to the curve $2x^{2}+y^{2}=33$ at $(2,5)$ is
$$
\begin{equation*}
y-5=\frac{5}{4}(x-2)\Leftrightarrow y=\frac{5}{4}x+\frac{5}{2},\tag{5}
\end{equation*}
$$
because the slope $m$ of the tangent line and the slope $m^{\prime }$ of the normal line are related by $mm^{\prime }=-1$.
ADDED. In a more general case, when we have a differentiable implicit
function $F(x,y)=0$, let $y=f(x)$ denote the function such that $F(x,f(x))\equiv 0\quad$ ($f(x)$ does not need to be explicitly known). If we differentiate both sides of $F(x,y)=0$ and apply the chain rule, we get the following total derivative with respect to $x$:
$$\frac{dF}{dx}=\frac{\partial F}{\partial x}+\frac{
\partial F}{\partial y}\frac{dy}{dx}\equiv 0.\tag{A}$$
Solving $(\mathrm{A})$ for $\frac{dy}{dx}$, gives us the following formula
$$\frac{dy}{dx}=-\frac{\partial F}{\partial x}/\frac{
\partial F}{\partial y}.\tag{B}$$
Best Answer
Your circle $\gamma$ is a level line of the function $$F(x,y):=x^2+y^2-2x-4y\ .$$ Therefore at any point $P=(x,y)\in\gamma$ the gradient $\nabla F(P)=(2x-2,2y-4)$ is orthogonal to the tangent at $P$. Since we want the points $P\in\gamma$ where the tangent is horizontal we want the points where the normal $\nabla F(P)$ is vertical, i.e., where $2x-2=0$, or $x=1$. Putting $x=1$ in the equation for $\gamma$ leads to the equation $$1+y^2-2-4y=-1$$ with the two solutions $y_1=0$ and $y_2=4$. It follows that there are two points where $\gamma$ has a horizontal tangent, namely the points $P_1=(1,0)$ and $P_2=(1,4)$.