When the curves meet, the \(r\) values and the \(\theta \) values must be both equal, and hence \begin {align*} a + 2 \cos \theta & = 2 + \cos 2\theta \\ a + 2 \cos \theta & = 2 + 2 \cos ^2 \theta - 1 \\ 2 \cos ^2 \theta - 2 \cos \theta + 1 - a & = 0, \end {align*}
as desired.
By differentiating with respect to theta, for the two curves to touch, we must have \begin {align*} \DiffOp {\theta } (a + 2 \cos \theta ) & = \DiffOp {\theta } (2 + \cos 2 \theta ) \\ - 2 \sin \theta & = - 2 \sin 2 \theta \\ \sin \theta & = \sin 2 \theta \\ \sin \theta & = 2 \sin \theta \cos \theta \\ \sin \theta (2 \cos \theta - 1) & = 0. \end {align*}
This means, either for the value of \(\sin \theta = 0\) it satisfies the first equation, or for the value of \(2 \cos \theta - 1 = 0\) it satisfies the first equation.
For the first case, we must have \(\cos \theta = \pm 1\), and hence \begin {align*} a & = 2 \cos ^2 \theta - 2 \cos \theta + 1 \\ & = 2 (\pm 1)^2 - 2 (\pm 1) + 1 \\ & = 3 \pm 2, \end {align*}
and so \(a = 1\) or \(a = 5\).
For the second case, we have \(\cos \theta = \frac {1}{2}\), and hence \begin {align*} a & = 2 \cos ^2 \theta - 2 \cos \theta + 1 \\ & = 2 \left (\frac {1}{2}\right )^2 - 2 \left (\frac {1}{2}\right ) + 1 \\ & = \frac {1}{2}, \end {align*}
as desired.
For the case where \(a = \frac {1}{2}\), the curves meet precisely for \(\cos \theta = \frac {1}{2}\) only, and hence \(\theta = \pm \frac {\pi }{3}\), which gives \(r = \frac {1}{2} + 1 = \frac {3}{2}\).
Both curves are symmetric about the initial line, since \(\cos \) is an even function.
When \(\theta = 0\), \(r_1 = a + 2 = \frac {5}{2}\), and \(r_2 = 2 + 1 = 3\).
For \(r_1\), since \(r \geq 0\), we must have \begin {align*} \frac {1}{2} + 2 \cos \theta & \geq 0 \\ \cos \theta & \geq - \frac {1}{4}, \end {align*}
which means it only exists for \[ -\arccos \left (-\frac {1}{4}\right ) \leq \theta \leq \arccos \left (-\frac {1}{4}\right ). \]
When \(\theta = \pm \frac {\pi }{2}\), \(r_1 = \frac {1}{2} + 2 \cos \pm \frac {\pi }{2} = \frac {1}{2}\).
For all values of \(\theta \), we must have \(r_2 \geq 0\). When \(\theta = \pi \), \(r_2 = 2 + 1 = 3\), and for \(\theta = \pm \frac {\pi }{2}\), \(r_1 = \frac {1}{2} + \cos \pm \frac {\pi }{2} = \frac {1}{2}\), \(r_2 = 2 + \cos \pm \pi = 1\).
Hence, the two curves are as follows. All coordinates are in \((r, \theta )\).
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\(a = 1\). For \(r_1\), since \(r \geq 0\), we must have \begin {align*} 1 + 2 \cos \theta & \geq 0 \\ \cos \theta & \geq -\frac {1}{2}, \end {align*}
which means \(-\frac {2}{3} \pi \leq \theta \leq \frac {2}{3}\pi \).
The two curves meet when \begin {align*} 2 \cos ^2 \theta - 2 \cos \theta & = 0 \\ \cos \theta (\cos \theta - 1) & = 0, \end {align*}
which is when \(\cos \theta = 0\) or \(\cos \theta = 1\).
For \(\cos \theta = 0\), this means \(\theta = \pm \frac {\pi }{2}\), and \(r = 1\). For this value of \(\theta \), the two curves cross.
For \(\cos \theta = 1\), this means \(\theta = 0\), and \(r = 3\). For this value of \(\theta \), the two curves touch.
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\(a = 5\). For \(r_1\), \(r \geq 0\) for all \(\theta \).
The two curves meet when \begin {align*} 2 \cos ^2 \theta - 2 \cos \theta & = 4 \\ \cos ^2 \theta - \cos \theta - 2 & = 0 \\ (\cos \theta - 2) (\cos \theta + 1) & = 0, \end {align*}
which is when \(\cos \theta = -1\), since \(\cos \theta \neq 2\).
For \(\cos \theta = -1\), this means \(\theta = \pi \), and \(r = 3\). For this value of \(\theta \), the two curves touch.
When \(\theta = 0\), \(r_1 = 5 + 2 = 7\), and \(r_2 = 2 + 1 = 3\). When \(\theta = \pm \frac {1}{2}\pi \), \(r_1 = 5 + 2 \cos \pm \frac {1}{2}\pi = 5\), \(r_2 = 2 + \cos \pm \pi = 1\).
