Find all $k$ such that the graph of parametric equations $\begin{align*} x &= 2+ 4\cos s, y= k-4\sin s, \end{align*}$ intersects the graph of the parametric equations $\begin{align*} x&=1+\cos t, y=-3+\sin t \end{align*}$ at only one point.
How would I find the constant $k$? I know that both parametric equations are circles, and that the second parametric circle is the circle with center $(1,-3)$ and radius $1$.
Best Answer
$(\cos t,\sin t)$ is the constant speed parametrization of the unit circle. Adding to the coordinates shifts its center, while multiplying these terms scales the circle and hence changes its radius. So as you already recognized, $(1+\cos t,â3+\sin t)$ is a circle or radius $1$ with center $(1,-3)$. The other, $(2+4\cos s,kâ4\sin s)$, is a circle with center $(2,k)$ and radius $4$.
The fact that you have $-4\sin s$ instead of $+4\sin s$ means that the parametrization is the other way round from the conventional one: you start at the righternmost point, but then go down (negative $y$ direction) instead of up. Doesn't change which circles this describes, only what parameter $s$ corresponds to which point on that circle.
So you have two circles, and you want to know how to choose the $y$ coordinate for the center of the second in such a way that the two circles have exactly one point in common, i.e. they touch. There are two ways the circles might touch: they might touch on the outside, or the smaller one might touch the larger one from the inside. In the former case, the distance between the centers has to be the sum of the radii, in the latter case it's the difference of the radii. So you have touching circles for
$$ \sqrt{(2-1)^2+(k+3)^2}\in\{4+1,4-1\}=\{5,3\} $$
To solve this, square the equation and you get
\begin{align*} (2-1)^2+(k+3)^2 &= 5^2 & (2-1)^2+(k+3)^2 &= 3^2 \\ k^2 + 6k + 9 &= 25 - 1 & k^2 + 6k + 9 &= 9-1 \\ k^2 + 6k - 15 &= 0 & k^2 + 6k + 1 &= 0 \\ k_{1,2} = \frac{-6\pm\sqrt{36+60}}{2} &= -3\pm2\sqrt6 & k_{3,4} = \frac{-6\pm\sqrt{36-4}}{2} &= -3\pm2\sqrt2 & \end{align*}