<s>[INST] Find the angle between two bivariate functions.. Find the angle between $x^2+y^2=8$ and $xy=4$ at the intersection points.
I have thought in find the angle between the tangent lines of each function at the intersection point, but i don´t know how to do it
 [/INST] Let the intersection point be $(a,b)$, so $a^2+b^2=8$  and $ab=4≠0$
$$\implies \frac{a^2+b^2}{ab}=\frac{4}{2}\implies (a-b)^2=0\implies a=b≠0$$
So,$a=±2$ as $ab=4$.
$$x^2+y^2=4\implies 2x+2y\frac{dy}{dx}=0\implies (\frac{dy}{dx})_{x=y}=-1$$
$$xy=4\implies x\frac{dy}{dx}+y=0\implies (\frac{dy}{dx})_{x=y}=-1 $$
So, at $(a,a)$ the gradients of the two given curves are same. 
We know if the angle between the two curves is $\theta$, then $\tan\theta=\frac{g_1-g_2}{1+g_1g_2}$ where $g_i$ s are the gradient of the intersecting curves.
So , the angle between the given two curves at $(a,a)$, i.e., at  $(2,2)$ and $(-2,-2)$, is
 $$\tan^{-1}\left(\frac{-1-(-1)}{1+(-1)(-1)}\right)=\tan^{-1}(0)=0 \text{ or }\pi$$
Alternatively, 
we are given $x^2+y^2=8$ and $xy=4$. From the latter, $x=\frac{4}{y}$
$$\left(\frac{4}{y}\right)^2+y^2=8\implies y^4-8y^2+16=0\implies (y^2-4)^2=0\implies y=±2$$
$$\tag 1 y=2\implies x=2$$ and $$\tag 2 y=-2\implies x=-2$$
So, $x=y$ at the points of intersection.
Observe that the equation of the common tangents are $x+y=a+a=2a$ i.e., $x+y=±4$.
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