Find locus with circle and intersection of lines 
$M(x_0,y_0), P(x,y)$
The equation of the circle is $x^2+y^2=25$.
MN is a chord in the circle perpendicular to the axis $x$. P is the intersection of BM and NA.
I found that 
$$MB: y = \frac{y_0}{x_0+5}x+\frac{5y_0}{x_0+5}$$
$$NA: y = \frac{y_0}{5-x_0}x-\frac{5y_0}{5-x_0}$$
I need to find the locus of the points P.
 A: With $O=(0,0)$ as circle center, let $\theta=\angle AOM$. Then the slope of line $BM$ is
\begin{equation}
m_1=\frac{\sin\theta}{1+\cos\theta}
\end{equation}
and the slope of line $NA$ is 
\begin{equation}
m_2=\frac{\sin\theta}{1-\cos\theta}
\end{equation}
So the point $P$ lies at the intersection of the lines
\begin{eqnarray}
BM:\quad y&=&\frac{\sin\theta}{1+\cos\theta}(x+5)\\
NA:\quad y&=&\frac{\sin\theta}{1-\cos\theta}(x-5)
\end{eqnarray}
which have point of intersection
\begin{equation}
P=\left(5\sec\theta,5\tan\theta\right)
\end{equation}
Since $\sec^2\theta-\tan^2\theta=1$ we have the hyperbola
\begin{equation}
\frac{x^2}{5^2}-\frac{y^2}{5^2}=1
\end{equation}
EDIT: Since the OP wants a solution not referencing trigonometry, I am adding the following additional solution:
First, give the point $M$ coordinates $(s,t)$ then find the coordinates of $P$ in terms of $s$ and $t$.  Since $(s,t)$ lies on the circle we know that
$$ s^2+t^2=25$$
and that, therefore
$$ 25-t^2=s^2\tag{1} $$
We will need equation $(1)$ below.
The slope of line $BM$ is $m_1=\dfrac{t}{s+5}$ and the slope of line $NA$ is $m_2=\dfrac{t}{5-s}$ so the equations of the two lines are
$$ BM:\quad y=\frac{t}{s+5}(x+5) $$
and 
$$ NA:\quad y=\frac{t}{5-s}(x-5) $$
The two lines intersect at $P=(x,y)$. Solving the two equations for $x$ and $y$ yields
$$ P=\left(\frac{25}{s},\frac{5t}{s}\right)$$
Then
\begin{eqnarray}
x^2-y^2&=&\frac{625-25t^2}{s^2}\\
&=&25\left(\frac{25-t^2}{s^2}\right) \text{ which, using equation } (1)\\
&=&25\cdot\frac{s^2}{s^2}\\
&=&25
\end{eqnarray}
Thus the equation of the locus of points $P$ is
$$  x^2-y^2=25$$

A: So I let the equation of the cord MN be $x = t$ where $t\in (-5,5)$. Then its is clear that the line BP has the form $y = \frac{\sqrt{25-t^2}}{5+t} x + c$. Substituting $(-5,0)$ I find that $c =  \frac{5 \sqrt{25-t^2}}{5+t}$. Similarly, we get the equation of the line NP: $y = \frac{\sqrt{25-t^2}}{5-t}[ x -5]$. To find the point P we set these two equations to be equal and calculate: $$\frac{\sqrt{25-t^2}}{5+t} [x+5] =  \frac{\sqrt{25-t^2}}{5-t} [x-5]$$ we can then cancel the $\sqrt{25-t^2}$ because $t\not\in\{-5,5\}$. Rearranging we get $x^2 -25 = 25-t^2$ and so when $t>0$ we have $x = \sqrt{50-t^2}$ and when $t< 0$ we have $x = -\sqrt{50-t^2}$. We can then find $y$ by substitution into one of our original equations.
