Abgle Between Curves at a Point

Suppose two curves meet at a point. We can find the angle between the curves by considering the gradients of the tangents to the two curves at the point.
Example:
$C_1 :x^2+y^2=25, \; C_2:y=x^2-5$
.
To find intersection points solve the simultaneous equations
$x^2+y^2=25$
(1)
$x^2-y=5$
(2)
(2)-(1) gives
$y^2+y=20 \rightarrow y^2+y-20=0 \rightarrow(y+5)(y-4)=0 \rightarrow y=-5, \; 4$
.
Take
$y=4$
then
$x^2=4+5=9 \rightarrow x=-3, \; 3$
.
$C_1:x^2+y^2=25$
is
$\frac{dy}{sx}- \frac{x}{y}$
so at
$(3,4)$
$m= \frac{dy}{dx} |_{(3,4)}=- \frac{3}{4}$
.
$C_2:y=x^2-5$
is
$\frac{dy}{dx}=2x$
so at
$(3,4)$
$m=\frac{dy}{dx} |_{(3,4)} = 6$
.
We can take tangent vectors to be
$\mathbf{u} = \begin{pmatrix}3\\ -4 \end{pmatrix}$
and
$\mathbf{v} = \begin{pmatrix}1\\ 6 \end{pmatrix}$
.
The angle between the curves is
$cos \theta = \frac{ \mathbf{u} \cdot \mathbf{v}}{ | \mathbf{u} | | \mathbf{v} | }= \frac{\begin{pmatrix}3\\ -4 \end{pmatrix} \cdot \begin{pmatrix}1\\ 6 \end{pmatrix}}{ \sqrt{3^2+(-4)^2} \sqrt{1^2+6^2}}= \frac{3 \times 1+ (-4) \times 6}{5 \sqrt{37}}= - \frac{21}{5 \sqrt{37}}$

Then
$\theta = cos^{-1}( - \frac{21}{5 \sqrt{37}})=133.67^o$
.