In this case, we have either \(y^2 + (x - 1)^2 = 1\) (giving a circle with radius \(1\) centred at \((1, 0)\)), or \(y^2 + (x + 1)^2 = 1\) (giving a circle with radius \(1\) centred at \((-1, 0)\)).
At \(y = k\), we have \begin {align*} [(x - 1)^2 + (k^2 - 1)] [(x + 1)^2 + (k^2 - 1)] & = \frac {1}{16} \\ (x - 1)^2 (x + 1)^2 + (k^2 - 1) [(x - 1)^2 + (x + 1)^2] + (k^2 - 1)^2 - \frac {1}{16} & = 0 \\ (x^2 - 1)^2 + 2 (k^2 - 1) (x^2 + 1) + (k^4 - 2k^2 + 1) - \frac {1}{16} & = 0 \\ x^4 - 2x^2 + 1 + 2 (k^2 - 1) x^2 + 2 (k^2 - 1) + (k^4 - 2k^2 + 1) - \frac {1}{16} & = 0 \\ x^4 + 2 (k^2 - 2) x^2 + k^4 - \frac {1}{16} & = 0, \end {align*}
as desired.
By completing the square, we have \begin {align*} (x^2 + (k^2 - 2))^2 + k^4 - \frac {1}{16} - (k^2 - 2)^2 & = 0 \\ (x^2 + (k^2 - 2))^2 & = \frac {65}{16} - 4k^2. \end {align*}
- If \(4k^2 > \frac {65}{16}\), i.e. \(k^2 > \frac {65}{64}\), the right-hand side is negative, so there will be no intersections.
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If \(4k^2 = \frac {65}{16}\), i.e. \(k^2 = \frac {65}{64}\), we have \[ x^2 + (k^2 - 2) = 0, \] and hence \[ x^2 = 2 - k^2 = 2 - \frac {65}{64} = \frac {63}{64}, \] giving \[ x = \pm \frac {3\sqrt {7}}{8}. \]
There will be two intersections.
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If \(4k^2 < \frac {65}{16}\), i.e. \(k^2 < \frac {65}{64}\), we have \[ x^2 + (k^2 - 2) = \pm \sqrt {\frac {65}{16} - 4k^2}, \] and hence \[ x^2 = 2 - k^2 \pm \sqrt {\frac {65}{16} - 4k^2}. \]
The case where \begin {align*} x^2 & = 2 - k^2 + \sqrt {\frac {65}{16} - 4k^2} \\ & > 2 - k^2 \\ & > 2 - \frac {65}{64} \\ & = \frac {63}{64} \\ & > 0 \end {align*}
always gives two solutions for \(x\).
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If \(2 - k^2 - \sqrt {\frac {65}{16} - 4k^2} < 0\), \begin {align*} 2 - k^2 - \sqrt {\frac {65}{16} - 4k^2} & < 0 \\ \sqrt {\frac {65}{16} - 4k^2} & > 2 - k^2 \\ \frac {65}{16} - 4k^2 & > k^4 - 4k^2 + 4 \\ k^4 & < \frac {1}{16} \\ k^2 & < \frac {1}{4}, \end {align*}
there are no solutions for the case where the minus sign is taken.
- If \(2 - k^2 - \sqrt {\frac {65}{16} - 4k^2} = 0\), \(k^2 = \frac {1}{4}\), the minus sign produce precisely one solution \(x = 0\), giving \(3\) intersections in total.
- If \(2 - k^2 - \sqrt {\frac {65}{16} - 4k^2} < 0\), \(k^2 > \frac {1}{4}\), the minus sign will produce two additional roots, hence giving \(4\) intersections in total.
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To summarise, the number of intersections with the line \(y = k\) for each positive value of \(k\) is \[ \text {number of intersections} = \begin {cases} 0, & k^2 > \frac {65}{64}, k > \frac {\sqrt {65}}{8}, \\ 2, & k^2 = \frac {65}{64}, k = \frac {\sqrt {65}}{8}, \\ 4, & \frac {1}{4} < k^2 < \frac {65}{64}, \frac {1}{2} < k < \frac {\sqrt {65}}{8}, \\ 3, & k^2 = \frac {1}{4}, k = \frac {1}{2}, \\ 2, & k^2 < \frac {1}{4}, 0 < k < \frac {1}{2}. \end {cases} \]
When the point on \(C_2\) has the greatest possible \(y\)-coordinate, the two points have \(x\)-coordinates \[ x = \pm \frac {3\sqrt {7}}{8}, \] and on \(C_1\) has \[ x = \pm 1. \]
Since \(3 \sqrt {7} = \sqrt {63} < \sqrt {64} = 8\), we must have \(\frac {3\sqrt {7}}{8} < 1\), meaning those on \(C_2\) are closer to the \(y\)-axis than those on \(C_1\).
If both are negative, then the distance from \((x, y)\) to \((1, 0)\) and \((-1, 0)\) are both less than \(1\). But this is not possible, since the distance from \((1, 0)\) to \((-1, 0)\) is \(2\), which means the sum of the distances from \((x, y)\) to those points has to be at least \(2\).
Therefore, since the product of those two terms are positive for \(C_2\), and they cannot be both negative, they must both be positive, and hence the distance from \((x, y)\) to \((1, 0)\) and \((-1, 0)\) are both more than \(1\), meaning all points on \(C_2\) lies outside the two circles that make up \(C_1\), which shows that \(C_2\) lies entirely outside \(C_1\).
\(C_2\) is symmetric about both the \(x\)-axis and the \(y\)-axis.
When \(x = 0\), \(y^4 = \frac {1}{16}\), and hence \(y = \pm \frac {1}{2}\).
When \(y = 0\), \(x^2 = 2 + \frac {\sqrt {65}}{16}\), and hence \(x = \pm \sqrt {2 + \frac {\sqrt {65}}{{4}}} = \pm \frac {1}{2} \sqrt {8 + \sqrt {65}}\).
Hence, the graph looks as follows.
