Locus of Center of a Circle while keeping the angle between the two external tangents (from the same point) constant

Question

if (D) is a fixed line and F is a fixed point that doesn’t belong to (D).

we have (C) a variable circle of center M passing through F and not cutting (D)

H is the orthogonal projection of M over (D). The tangents drawn at H to (C) cut it at T and T’ such that THT’ = 2X remains constant.

how can I find the locus of M?

i’ve been trying to figure it out for a bit now but it doesn’t seem to click in my head yet

Answer 1

Choose a coordinate system such that line $D$ is the x-axis and $F=(0,f)$ is on the y-axis. If $M=(x,y)$, from $TM^2=FM^2$ you can then derive the equation of the locus, as $TM=y\sin X$: $$ (y\sin X)^2=x^2+(y-f)^2. $$

EDIT.

For a synthetic solution, just observe that:

$$ {FM\over HM}={TM\over HM}=\sin X. $$ Hence point $M$ belongs to an ellipse of eccentricity $\sin X$, having $F$ as focus and line $D$ as directrix.