Find Angle Between Two Curves at Point of Intersection Find angle between these two curves at point of intersection : 
$$K_1: x^2y^2 + y^4 = 1$$ and 
$$K_2 : x^2 + y^2 = 4 $$ 
Thanks!

Problem
Given the equations
$$
 \color{blue}{x^{2} + y^{2} = 4}
\tag{1}
$$
and
$$
 x^{2}y^{2} + y^{4} = 1
\tag{2}
$$
find the angles between the curves at the points of intersection.

Symmetry analysis
Each term in both equations has a definite parity; this invites investigation. 
The $x^{2}$ terms are invariant under the transformation $x\to-x$ because
$$
 (-x)^{2} = x^{2}
$$
A similar argument holds for both $y^{2}$ and $y^{4}$.
From this we see invariance in form when both figures are reflected through each of the $x-$axis and the $y-$axis. Therefore, both figures are invariant under reflection through the origin.  These ideas are demonstrated in the sequence of images below.

Conclusion: we only need to solve the problem at one intersection point. 
Find intersections
Rewrite $(2)$:
$$
 y^{2} \left( x^{2} + y^{2} \right) = 1
\tag{2a}
$$
and inserting $\color{blue}{(1)}$ we have
$$
 y^{2}\left( 4 \right) = 1
$$
or
$$
  y = \frac{1}{2}
\tag{3}
$$
Insert $(3)$ into $(1)$ to find
$$
 x = \frac{\sqrt{15}}{2}.
$$
by the symmetry analysis, the locus of intersecting points is
$$
 \left\{
\frac{1}{2} \left( \sqrt{15},   1 \right ),
\frac{1}{2} \left( \sqrt{15},  -1 \right ),
\frac{1}{2} \left( -\sqrt{15},  1 \right ),
\frac{1}{2} \left( -\sqrt{15}, -1 \right )
 \right\}
$$
We will focus on the point $p = \frac{1}{2} \left( \sqrt{15},   1 \right )$.
Compute angles
Here are the slopes at the intersection point:

The slope calculation is detailed in the answer by @Manuel Guillen. Now to relate the slopes to angles.
Given the line
$$
 y = mx + b
$$
the slope is related to the angle by
$$
 m = \tan \theta
$$
The figure below shows the intersection angle as an arc between the tangent lines. The arc goes from 
$$
-\tan ^{-1}\left(\frac{\sqrt{15}}{17}\right) \le \theta \le
\pi -\tan ^{-1}\left(\sqrt{15}\right) 
$$

The difference $\Delta$, is the angle of intersection:

$$
\Delta = \pi +\tan ^{-1}\left(\frac{\sqrt{15}}{17}\right)-\tan ^{-1}\left(\sqrt{15}\right) =
\pi -\cot ^{-1}\left(\frac{2}{\sqrt{15}}\right) 
\approx 117^{\circ}
$$

