Given a point, a line and a circle; how can I find a point on the line having the following property? First of all: I searched with different search engines and on StackExchange, but did not find a solution to my problem. This may be because its specific description is kind of long. For the same reason, I had to choose an unspecific title to this question due to the character limit.
Now, here is my problem:
Given

*

*A point $D$

*A line $l$

*A circle $K$ (with center $C$ and radius $r$)

all in $\mathbb{R}^2$

I want to find a point $G$ such that:

*

*$G$ is incident to $l$

*Let the intersection of the line $GC$ with $K$ be $H$. It should hold that $G$ has the same distance to $D$ as to $H$, i.e. $\left\lVert\vec{GD}\right\rVert = \left\lVert\vec{GH}\right\rVert$

So far I tried to find some property that this point has that allows me to construct it, but haven't found one yet. I am also very inexperienced and have little to no knowledge in geometry, which might be why this is hard to me.
Edit (clarifications):

*

*$D$ and $C$ may be assumed to lie on the same side of $l$ (I guess this can even be assumed w.l.o.g., as we could otherwise just mirror the point $D$).

*$H$ is the intersection point that is further away from $G$ than $C$.

*$D$ can be assumed to be exterior $K$.

*$l$ can be assumed to intersect $K$ at two different points and to not pass through $C$.

*I preferably look for geometric solutions or algebraic solutions in closed form.

 A: *

*Reflect $D$ in the line $l$, denote the image $D^*$.


*Draw the line $DD^*$.


*Pick an arbitrary, but convenient, point $E$ on the line $l$ and draw the circle with center $E$ and radius $ED = ED^*$. It is very helpful if this circle $K(E, ED)$ intersects the the original one $K$ at two points.


*Draw the radical axis of the two circles $K(E, ED)$ and $K$. If the two circle intersect, then the radical axis is the line that passes through both intersection points of the two circles.


*Denote by $F$ the intersection point between the radical axis and the line $DD^*$


*Construct the line through point $F$ and tangent to the circle $K$. There are two of them, so you have two solutions.


*Denote by $T$ the point of tangency between the line from point 5 and the circle $K$.


*The circumcircle of triangle $\Delta \, TDD^*$ is the circle you are looking for (there are two solutions).
A: Sketch for an algebraic solution.
Translate, rotate and scale your picture so that the circle is the unit circle, the line $\ell$ is horizontal at known height $y=h$. Then $D$ will have known coordinates $(c,d)$. The distance from $D$ to $\ell$ is $d-h$. Suppose $G=(x,h)$. Then the line from $G$ through the center of the circle has meets the circle at the two points $\theta$ for which $b/a = \tan(\theta)$. Moreover, $G = ((\tan \theta)/h, h)$.
Equate the squares of the two distances you care about:
$$
(c- ((\tan \theta) / h)^2 + (d- h)^2= (a-\cos \theta)^2 + ( b - \sin \theta)^2
$$
That's sufficient information to solve for the angle $\theta$. You may encounter algebraic problems that will reflect the fact that the ambiguities in the picture suggest cases with no solutions and multiple solutions.
