Find the angle of intersection at all points where the graphs meet between $(x-2)^2 + (y+1)^2 = 16$ and $y = \frac{3}{2}x$.
I solved this by finding the easier point of intersection $(2,3)$, which was conveniently at the vertex of the circle thus the tangent line is simply $y = 3$. I then used elementary trig to deduce that the angle of intersection is $\sin^{-1}(\frac{3}{\sqrt{13}}) \approx .9827$. My argument is that this angle should be the same for the other intersection as well. I carried on finding it anyways, the intersection occurs at $(\frac{-22}{13},\frac{-33}{13})$ and I found the angle here to be approximately the same, however it took much longer to find since the numbers here are nastier.
I am doing this problem because I am studying for an exam that covers a lot of old calc material that I am not fresh on. It has been a while. Speed is very important for this exam; my question is: does anyone see a much faster or more efficient route to solving this? I was thinking maybe the cross product could be used? And if the idea that the angles are the same is correct could someone prod me towards justifying that?
Thank you.
The other "angle of intersection" is exactly the same for two reasons:-
1) That straight line passes through the origin; and
2) The two angles are "vertically opposite angles".
Hence, no further computation is necessary.
Added:
$\alpha = \beta$ … [vertically opposite angles]
$\beta = \gamma$ … [corresponding angles]
In addition, $\alpha = \delta$ ... [alternate angles]
$\delta = \theta$ ... [tangent properties]