Problem source

\begin{questionparts}
\item The circle $x^2 + (y-a)^2 = r^2$ touches the parabola $2ky = x^2$, where $k > 0$, tangentially at two points. Show that $r^2 = k(2a - k)$.
Show further that if $r^2 = k(2a - k)$ and $a > k > 0$, then the circle $x^2 + (y-a)^2 = r^2$ touches the parabola $2ky = x^2$ tangentially at two points.
\item The lines $y = c \pm x$ are tangents to the circle $x^2 + (y-a)^2 = r^2$. Find $r^2$, and the coordinates of the points of contact, in terms of $a$ and $c$.
\item $C_1$ and $C_2$ are circles with equations $x^2 + (y-a_1)^2 = r_1^2$ and $x^2 + (y-a_2)^2 = r_2^2$ respectively, where $a_1 \neq a_2$ and $r_1 \neq r_2$.
Each circle touches the parabola $2ky = x^2$ tangentially at two points and the lines $y = c \pm x$ are tangents to both circles.
\begin{enumerate}
\item Show that $a_1 + a_2 = 2c + 4k$ and that $a_1^2 + a_2^2 = 2c^2 + 16kc + 12k^2$.
\item The circle $x^2 + (y-d)^2 = p^2$ passes through the four points of tangency of the lines $y = c \pm x$ to the two circles, $C_1$ and $C_2$. Find $d$ and $p^2$ in terms of $k$ and $c$.
\item Show that the circle $x^2 + (y-d)^2 = p^2$ also touches the parabola $2ky = x^2$ tangentially at two points.
\end{enumerate}
\end{questionparts}

Solution source

\begin{questionparts}
\item \begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){2*(#1)*((#1)^2 - 5)/((#1)^2-4)};
    \def\xl{-2};
    \def\xu{2};
    \def\yl{-0.2};
    \def\yu{3.8};
    
    % Calculate scaling factors to make the plot square
    \pgfmathsetmacro{\xrange}{\xu-\xl}
    \pgfmathsetmacro{\yrange}{\yu-\yl}
    \pgfmathsetmacro{\xscale}{10/\xrange}
    \pgfmathsetmacro{\yscale}{10/\yrange}
    
    % Define the styles for the axes and grid
    \tikzset{
        axis/.style={very thick, ->},
        grid/.style={thin, gray!30},
        x=\xscale cm,
        y=\yscale cm
    }
    
    % Define the bounding region with clip
    \begin{scope}
        % You can modify these values to change your plotting region
        \clip (\xl,\yl) rectangle (\xu,\yu);
        
        % Draw a grid (optional)
        % \draw[grid] (-5,-3) grid (5,3);
        
        \draw[thick, blue, smooth, domain=0:360, samples=100] 
            plot ({cos(\x)}, {sin(\x)+1.5});
        \draw[thick, blue, smooth, domain=-2:2, samples=100] 
            plot ({\x}, {\x*\x/(3-sqrt(5))});
        
        % \draw[thick, red, dashed, domain=-5:5, samples=100] 
        %     plot (\x, {2*\x*\x*\x-5*\x});
        % \draw[thick, blue, smooth, domain=-5:5, samples=100] 
        %     plot (\x, {2*min(\x*\x, \x*\x*\x) - 5*\x});  
        % \draw[thick, red, dashed] (-2, \yl) -- (-2, \yu) node [below] {$x = -2$};
        % \draw[thick, red, dashed] (2, \yl) -- (2, \yu) node [below] {$x = 2$};
        % \draw[thick, red, dashed] (-5, -10) -- (5, 10) node [below left] {$y = 2x$};
    \end{scope}
    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right] {$x$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
    
    \end{tikzpicture}
\end{center}

By symmetry we can observe that the parabola and circle will intersect $0, 1$ (at the base), $2, 4$ times. So setting up our system of equations we have:

\begin{align*}
&&& \begin{cases} x^2 + (y-a)^2 &= r^2 \\
2ky &= x^2 \end{cases} \\
\Rightarrow && r^2 &=  x^2 + \left (\frac{x^2}{2k} - a \right )^2  \\
\Rightarrow &&r^2 &= x^2 + a^2 - \frac{ax^2}{k} + \frac{x^4}{4k^2} \\
\Rightarrow &&0 &= \frac{1}{4k^2} x^4 + \left ( 1 - \frac{a}{k} \right) x^2 + a^2 - r^2 \\
\Rightarrow && \Delta &= \left ( 1 - \frac{a}{k} \right)^2-4 \cdot \frac{1}{4k^2} (a^2 - r^2) \\
&&&= 1 - \frac{2a}{k} + \frac{a^2}{k^2} - \frac{a^2}{k^2} + \frac{r^2}{k^2} \\
&&&= \frac{k^2-2ka+r^2}{k^2}
\end{align*}

Since there will be (at most) two solutions if $\Delta = 0$ we must have if the circle and parabola are tangent $r^2 - 2ka + k^2 = 0 \Rightarrow r^2 = k(2a-k)$. So long as there is a solution $x^2 > 0$ there will be two tangent points, so if $-\left(1 - \frac{a}{k}\right) > 0$ or $a > k > 0$

\item Since $y = c \pm x$ are tangent to the circle with radius $r$ and centre $(0,a)$ we have the following equations:

\begin{align*}
&&& \begin{cases} x^2 + (y-a)^2 &= r^2 \\
c \pm x &= y \end{cases} \\
\Rightarrow && r^2 &= x^2 + (c -a\pm x)^2  \\
&&&= 2x^2+(c-a)^2 \pm 2x(c-a) \\
\Rightarrow && \Delta &= 4(c-a)^2 -4 \cdot 2 \left ( (c-a)^2 -r^2 \right)\\ 
&&&= 8r^2-4(c-a)^2 \\
\Rightarrow && x &= \frac{\mp 2(c-a) \pm \sqrt{\Delta}}{4} \\
&&&= \mp \frac12 (c-a) \\
&& y &= \pm \frac12 (c+a) \\
&& (x,y) &= \left (\frac12 (c-a), \frac12 (c+a)\right), \left (-\frac12 (c-a), -\frac12 (c+a)\right)
\end{align*}


\item \begin{enumerate}
\item 
\end{enumerate}

\end{questionparts}