The graph of the parametric equations \begin{align*} x&=\cos t,\\ y&=\sin t,

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.

\( \text{circle 1:} \\ \begin{array}{|rcll|} \hline x &=& \cos(t) \\ y &=& \sin(t) \\ \hline x^2+y^2 &=& \cos^2(t) + \sin^2(t) \\ \mathbf{x^2+y^2} & \mathbf{=} & \mathbf{1} \\ \hline \end{array}\)

\(\text{circle 2:} \\ \begin{array}{|rcll|} \hline x &=& 2+4\cos(t) \quad \text{ or } \quad 4\cos(t) = x-2 \\ y &=& 3+4\sin(t) \quad \text{ or } \quad 4\sin(t) = y-3 \\ \hline x^2+y^2 &=& \Big( 2+4\cos(t) \Big)^2 + \Big( 3+4\sin(t) \Big)^2 \\ x^2+y^2 &=& 4+16\cos(t)+16\cos^2(t) + 9 + 24\sin(t) + 16\sin^2(t) \\ x^2+y^2 &=& 13+16\cos(t)+ 24\sin(t) + 16\Big(\cos^2(t) + \sin^2(t) \Big) \\ x^2+y^2 &=& 13+16\cos(t)+ 24\sin(t) + 16 \\ x^2+y^2 &=& 29+16\cos(t)+ 24\sin(t) \\ x^2+y^2 &=& 29+4\cdot4\cos(t)+ 6\cdot 4\sin(t) \\ x^2+y^2 &=& 29+4\cdot (x-2)+ 6\cdot (y-3) \\ x^2+y^2 &=& 29+4x-8+6y-18 \\ \mathbf{x^2+y^2} & \mathbf{=} & \mathbf{3+4x +6y} \\ \hline \end{array} \)

\(\text{intersection between the two circles:} \\ \begin{array}{|rcll|} \hline \text{circle 2: }~\mathbf{x^2+y^2} & \mathbf{=} & \mathbf{3+4x +6y} \quad &| \quad \text{circle 1: }~ \mathbf{x^2+y^2= 1} \\ 1 &=& 3+4x +6y \\ 6y+4x+3 &=& 1 \\ 6y &=& -4x-3 +1 \\ 6y &=& -4x-2 \\\\ y &=& \dfrac{-4x-2}{6} \\\\ y &=& -\dfrac{4}{6}x -\dfrac{2}{6} \\\\ y &=& -\dfrac{2}{3}x -\dfrac{1}{3} \\ \hline \end{array} \)

\(\text{the line between these two points:} \\ \begin{array}{|rcll|} \hline \mathbf{ y } & \mathbf{=} & \mathbf{ \underbrace{-\dfrac{2}{3}}_{\text{slope}}x -\dfrac{1}{3} } \\ \hline \end{array} \)

The slope of the line between these two points is \(\mathbf{-\dfrac{2}{3}}\).