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.

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.
