Locus of points generates several very different curves. Closed form? Consider, for the sake of simplicity, a circle $C$ centered at he origin with radius $a$. Let $F=(h,k)$ be a point not necessarily inside the circle. Let $M=(a\cos\theta,a\sin\theta)$ be a point in the circle, and $L$ the line that pases through $M$ and $F$. Let $L_1$ be the line passing through the midpoint of $MO$ where $O$ is the origin, perpendicular to $MO$. Then the curve sought is the locus of points of the point of intersection of $L$ and $L_1$. 
You can check that these two lines can be written as
$$y =  - \frac{1}{{\tan (t)}}\left( {x - \frac{{a\cos \left( t \right)}}{2}} \right) + \frac{{a{\text{ }}\sin (t)}}{2}{\text{ }}$$
and
$$y = \frac{{a\sin (t) - k}}{{a\cos (t) - h}}{\text{ }}(x - h){\text{ }} + k$$
This, what we want to is express $x$ and $y$ in terms of $t$ and the curve will be $(x(t),y(t))$. Now, although the plotting soft produces the right curves, I have failed $4$ times already to solve for $x$ and $y$. I am getting
$$x(t) = \frac{{2a\cot t\cos t  + 2a\sin t  + {\text{2}}f\left( t  \right)h - 2k}}{{{\text{ 2}}f\left( t  \right) + 2\cot t }}{\text{ }}$$
and
$$y(t)=f(t)(x(t)-h)+k$$
where $${\text{ }}\frac{{a\sin (t) - k}}{{a\cos (t) - h}} = f\left( t \right)$$
but this fails to give the correct parametrization.
Could you help to find a closed expression in terms of $t$ and verify it is correct?
These are pics of the curves generated. If the focus is inside the cirlce of radius $a/2$ we get a deformed circle, if it is in the half outside, we get a closed intersecting curve, if it is exaclty on the circle we get the Foluim of Descartes/Trisectrix of Maclaurin, if it is right in the origin, another circle, if it is exactly halfway, a cardioid, and if it is outside the circle, a distorted Foluim of Descartes/Trisectrix of Maclaurin with a part of an hiperbola.






 A: By scaling and rotation, we can assume the radius of the circle is $1$ and that $F=(f,0)$. The situation described above is diagrammed below. $P$ is the intersection of $L$ and $L_1$, $B$ is the midpoint of $\overline{OA}$, and $A$ is the point on $\overline{OM}$ so that $\overline{FA}\perp\overline{OA}$. By assumption, $|\overline{OM}|=1$, and by definition, let $f=|\overline{OF}|$.
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Note that $\angle AMF=\angle BMP$ since they are vertical angles. Since $\triangle AMF$ and $\triangle BMP$ are right triangles that share an acute angle, they are similar. Therefore,
$$
(P-M)|\overline{MA}|=(M-F)|\overline{MB}|\tag{1}
$$
Combining $(1)$ with $|\overline{MA}|=f\cos(\theta)-1$ and $|\overline{MB}|=\frac12$ yields
$$
P=\frac{\left(f\cos(\theta)-\frac12\right)M-\frac12F}{f\cos(\theta)-1}\tag{2}
$$
Using $(2)$ with $M=(\cos(\theta),\sin(\theta))$ and $F=(f,0)$ gives
$$
P=\frac{\frac12\left(f\cos(2\theta)-\cos(\theta),f\sin(2\theta)-\sin(\theta)\right)}{f\cos(\theta)-1}\tag{3}
$$
Animation
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A: From Maple,
$$x =\frac{a}{2} \frac{2k\sin(t)\cos(t)-h+2h\cos(t)^2-a\cos(t)}{(\sin(t)k+h\cos(t)-a)}$$ 
$$y =\frac{a}{2} \frac{2h\sin(t)\cos(t)-a\sin(t)-2k\cos(t)^2+k}{(\sin(t)k+h\cos(t)-a)}$$
