How does $\angle ACB$ change as the point $C$ changes?

Question

Let $A = (-4,0)$, $B = (4,0)$, $C = (x,y)$.

How would you determine how angle $\angle ACB$ changes in relation to $x$ and $y$? Must a further variable be introduced? If so, will knowing the distance between $C$ and the origin help? I've been puzzling at this for about a week now but cannot come to anything, so I assume there is a gap in my knowledge of such analytical geometry. Assuming that this is done through differential calculus in relation to geometry, once an equation is derived, it should be easy to find local minima and maxima.

Answer 1

If $\theta = \angle ACB$, then $AC = (x + 4, y)$ and $BC = (x - 4, y)$, so \begin{align*} \cos\theta &= \frac{AC \cdot BC}{\|AC\|\, \|BC\|} \\ &= \frac{x^{2} - 16 + y^{2}}{\sqrt{(x + 4)^{2} + y^{2}}\, \sqrt{(x - 4)^{2} + y^{2}}} \\[12pt] &= \sqrt{\frac{(x^{2} - 16 + y^{2})^{2}}{[(x + 4)^{2} + y^{2}][(x - 4)^{2} + y^{2}]}} \\[12pt] &= \sqrt{\frac{(x^{2} - 16)^{2} + 2y^{2}(x^{2} - 16) + y^{4}}{(x^{2} - 16)^{2} + 2y^{2}(x^{2} + 16) + y^{4}}}. \end{align*}

Answer 2

You can use dot product of the two vectors $\vec{CA}.\vec{CB}$, which is defined as $\vec{CA}.\vec{CB}=|\vec{CA}||\vec{CB}| \cos \angle{ACB}$, other expression for the dot product is $\vec{CA}.\vec{CB}=x_1y_1+x_2y_2$ if vector $\vec{CA}$ and vector $\vec{CB}$ are represented as cooridantes $(x_1,y_1)$ and $(x_2, y_2)$. You can use these two to derive the size of the cosine, which is enough to determine the angle.