The graph of the parametric equations \begin{align*} x&=\cos t,\\ y&=\sin t \end{align*} meets the graph of the parametric equations \begin{align*} x &= 2+ 4\cos s,\\ y &= 3+4\sin s \end{align*} at two points. Find the slope of the line between these two points.
I graphed it out on Desmos, but it wouldn't give me the points.
How should I solve?
The parametrization of \begin{align*} x&=\cos t \hspace{4mm}\mbox{ and } \hspace{4mm} y=\sin t \end{align*} is the circle $x^2+y^2=1$ of radius $1$ and the parametrization of \begin{align*} x &= 2+ 4\cos s \hspace{4mm}\mbox{ and } \hspace{4mm} y = 3+4\sin s \end{align*} is the circle $(x-2)^2+(y-3)^2=16$ of radius $4$. Expand $(x-2)^2+(y-3)^2=16$ to get $x^2 - 4 x +13 + y^2 - 6 y =16$. Then $$ x^2+y^2-(x^2 - 4 x +13 + y^2 - 6 y )= \color{green}{-13 + 4 x + 6 y}=1-16\color{green}{=-15}. $$ Since $3y=-2x-1$, we have $$ y=-\frac{2}{3}x-\frac{1}{3}. $$ Thus the slope between the two points at the intersection of the two circles is $$ -\frac{2}{3}. $$