Problem source

\begin{questionparts}
\item Sketch the curve $C_1$ with equation
\[ \left(y^2 + (x-1)^2 - 1\right)\left(y^2 + (x+1)^2 - 1\right) = 0. \]
\item Consider the curve $C_2$ with equation
\[ \left(y^2 + (x-1)^2 - 1\right)\left(y^2 + (x+1)^2 - 1\right) = \tfrac{1}{16}. \]
\begin{enumerate}
\item Show that the line $y = k$ meets the curve $C_2$ at points for which
\[ x^4 + 2(k^2 - 2)x^2 + \left(k^4 - \tfrac{1}{16}\right) = 0. \]
Hence determine the number of intersections between curve $C_2$ and the line $y = k$ for each positive value of $k$.
\item Determine whether the points on curve $C_2$ with the greatest possible $y$-coordinate are further from, or closer to, the $y$-axis than those on curve $C_1$.
\item Show that it is not possible for both $y^2 + (x-1)^2 - 1$ and $y^2 + (x+1)^2 - 1$ to be negative, and deduce that curve $C_2$ lies entirely outside curve $C_1$.
\item Sketch the curves $C_1$ and $C_2$ on the same axes.
\end{enumerate}
\end{questionparts}

Solution source

\begin{questionparts}
\item $\,$ 
\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){exp(1/(#1)*ln(#1))};
    \def\xl{-2.6}; 
    \def\xu{2.6};
    \def\yl{-2.6}; 
    \def\yu{2.6};
    
    % 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 reusable styles to keep code clean
    \tikzset{
        x=\xscale cm, y=\yscale cm,
        axis/.style={thick, draw=black!80, -{Stealth[scale=1.2]}},
        grid/.style={thin, dashed, gray!30},
        curveA/.style={very thick, color=cyan!70!black, smooth},
        curveB/.style={very thick, color=orange!90!black, smooth},
        curveC/.style={very thick, color=green!90!black, smooth},
        curveBlack/.style={very thick, color=black, smooth},
        dot/.style={circle, fill=black, inner sep=1.2pt},
        labelbox/.style={fill=white, inner sep=2pt, rounded corners=2pt} % Protects text from lines
    }
    
    % Draw background grid
    \draw[grid] (\xl,\yl) (\xu,\yu);
    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right, black] {$x$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above, black] {$y$};
    
    % Define the bounding region with clip
    \begin{scope}
        \clip (\xl,\yl) rectangle (\xu,\yu);

        \draw[curveA] (1,0) circle (1);
        \draw[curveA] (-1,0) circle (1);


    \end{scope}
    
    % Annotate Function Names
    % \node[curveB, labelbox] at (1.85, -1.1) {$x = \frac{4y^2+1}{3}$};
    

    \end{tikzpicture}
\end{center}

\item \begin{enumerate}
\item Suppose $y=k$ meets the curve $C_2$ then
\begin{align*}
&& \tfrac1{16} &= (k^2+(x-1)^2-1)(k^2+(x+1)^2-1) \\
&&&= (k^2+x^2-2x)(k^2+x^2+2x) \\
&&&= k^4+2k^2x^2+x^4-4x^2 \\
&&&= x^4+(2k^2-4)x^2+k^4 \\
\Rightarrow && 0 &= x^4+(2k^2-4)x^2+(k^4-\tfrac1{16}) \\
\\
&& \Delta &= 4(k^2-2)^2 - 4 \cdot 1 \cdot (k^4-\tfrac1{16}) \\
&&&= 4(k^4-4k^2+4 - k^4 +\tfrac{1}{16}) \\
&&&= 16(1+\tfrac{1}{64} - k^2) \\
&&&= 16(\tfrac{65}{64} - k^2)
\end{align*}
Therefore if $|k| < \frac{\sqrt{65}}{8}$ there are $4$ intersections. If $|k| =  \frac{\sqrt{65}}{8}$ there are $2$ intersections, otherwise there are $0$.
\item The greatest possible $y$ value is $ \frac{\sqrt{65}}{8}$ and at this point $x^2 = \frac{2(2-\frac{65}{64}}{2} =1 - \frac{1}{64} < 1$ so they are close to the $y$-axis.

\item The regions where $y^2+(x-1)^2-1 < 0$ and $y^2+(x+1)^2-1 < 0$ is the interior of the two circles from the first part. However, since they don't overlap they can never both be negative. Therefore  in our equation both are positive and therefore $C_2$ is entirely outside $C_1$

\item $\,$ 

\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){exp(1/(#1)*ln(#1))};
    \def\xl{-2.6}; 
    \def\xu{2.6};
    \def\yl{-2.6}; 
    \def\yu{2.6};
    
    % 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 reusable styles to keep code clean
    \tikzset{
        x=\xscale cm, y=\yscale cm,
        axis/.style={thick, draw=black!80, -{Stealth[scale=1.2]}},
        grid/.style={thin, dashed, gray!30},
        curveA/.style={very thick, color=cyan!70!black, smooth},
        curveB/.style={very thick, color=orange!90!black, smooth},
        curveC/.style={very thick, color=green!90!black, smooth},
        curveBlack/.style={very thick, color=black, smooth},
        dot/.style={circle, fill=black, inner sep=1.2pt},
        labelbox/.style={fill=white, inner sep=2pt, rounded corners=2pt} % Protects text from lines
    }
    
    % Draw background grid
    \draw[grid] (\xl,\yl) (\xu,\yu);
    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right, black] {$x$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above, black] {$y$};
    
    % Define the bounding region with clip
    \begin{scope}
        \clip (\xl,\yl) rectangle (\xu,\yu);

        \draw[curveA] (1,0) circle (1);
        \draw[curveA] (-1,0) circle (1);

        \draw[curveB, domain=-2.01:2.01, samples=400]
            plot ({\x},  {sqrt(sqrt(4*(\x)^2 + 1/6) - (\x)^2)});
        \draw[curveB, domain=-2.01:2.01, samples=400]
            plot ({\x},  {-sqrt(sqrt(4*(\x)^2 + 1/6) - (\x)^2)});

    \end{scope}
    
    % Annotate Function Names
    % \node[curveB, labelbox] at (1.85, -1.1) {$x = \frac{4y^2+1}{3}$};
    

    \end{tikzpicture}
\end{center}

\end{enumerate}
\end{questionparts}