The angle at which a circle and a hyperbola intersect? 
$x^2 - 2y^2 = 4$
  $ (x-3)^2 + y^2 = 25 $

How do you calculate the angle at which a circle and a hyperbola intersect?
If I express $y^2$ from the first equation and apply it to the second equation, I get the following:
$y^2 = -2 + \frac{x^2}{2}$
$(x-3)^2 + -2 + \frac{x^2}{2} = 25$ ... $x^2 - 4x - 12 = 0$
$x_1 = 6, x_2 = -2 \implies y_1 = 4, y_{1'} = -4, y_2 = 0$  
Now, for the points $(6,4)$ I could calculate the line equation which intersects the circle and the hyperbola: $(6-3)(x-3) + (4-0)(y-0) = 25 \implies y = -\frac{3x}{4} + \frac{17}{2}$  
I calculated this because I thought I could apply the formula $\tan\phi = |\frac{k_2 - k_1}{1 + k_1k_2}|$

HINT:  
$(1):$ From the article $151$ of The elements of coordinate geometry (Loney),  the gradient $(m_1)$ of $x^2+y^2+2gx+2fy+c=0$ is $=-\frac{x_1+g}{y_1+f}$ where $(x_1,y_1)$ is the given point on the circle
and from the Article $305,262$  the gradient $(m_2)$ of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $\frac{b^2\cdot x_2}{a^2\cdot y_2}$ where $(x_2,y_2)$ is the given point on the hyperbola
$(2):$ Alternatively, find the gradients of each curve at $(6,4)$ and at $(-2,0)$ applying first order derivative
The acute angle between the curves will be $$\arctan\left|\frac{m_1-m_2}{1+m_1m_2} \right|$$  
