# Find the intersection of two parametric equations at one point

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 $1$.

• you want to chose $k$, so the two circles touch each other. This requires $r_1+r_2=d$ where $d$ is the distance of the two centers and $r_1,r_2$ are the radii – Gregor de Cillia Feb 3 '15 at 19:40
• @GregordeCillia: You mean that $d$ is the distance between the two centers? I think that the first parametric equation is a circle with radius 4. So $r_1 = 4, r_2 = 1$, so $d = 5$? – Mathy Person Feb 3 '15 at 19:41
• Yes, now you have to find $k$ such that $\|(2,k)^T-(1,-3)^T\|=5$ – Gregor de Cillia Feb 3 '15 at 19:43
• @GregordeCillia: How would that work (subtracting points from each other)? Would it be: $(x_1, y_1) - (x_2, y_2) = (x_1-x_2,y_1-y_2)$? What is $T$? – Mathy Person Feb 3 '15 at 19:44
• @GregordeCillia: What is $^T$? – Mathy Person Feb 3 '15 at 20:29

$(\cos t,\sin t)$ 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, $(1+\cos t,−3+\sin t)$ is a circle or radius $1$ with center $(1,-3)$. The other, $(2+4\cos s,k−4\sin s)$, 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*} 