Find the location of point M where $MA-MB=2L$ , $Given length$

Question

$C(O,AB)$ Semicircle $AB=r$ Radius

Find the location of point M where $MA-MB=2L$ , $L=\operatorname{Given length}$

My try :

Let $\angle BAM=\theta $ so $MA≥MB$ where $\theta \in [0,\frac{π}{4}]$

Now ,

$$MA-MB=\cos \theta -\sin \theta =\frac{L}{r}$$

$$\cos (\theta +\frac{π}{4})=\frac{L}{2\sqrt{2}}$$

But I don't know how i complete ?

Answer 1

There is not one such point $m$, but whole set of them.

In fact they constitue whole (right) branch of hyperbola with $2e = AB =2r$ and $2a=2L$. So if line $AB$ is an $x$ axsis and $O$ origin then the equation of this hyperbola is $${x^2\over L^2}-{y^2\over b^2}=1$$ where $b^2 = r^2-L^2$.