https://i.stack.imgur.com/EEVNU.jpg
Please refer to link for image; given circle at center A, and line h-D, construct a circle tangent to the circle and tangent to line at point D. I have already done another problem similar to it but instead of point D, point B is given instead.
I want a theorem-based/backed approach. Use circle and lines for construction, and avoid parabolas, hyperbolas, and the like.
I'm going to pretend that the given line through $D$ is horizontal to make describing things simpler, even though it's not quite drawn that way.
The circle-center must be on the perpendicular $\ell$ to the horizontal at $D$, which you can construct. Extend the line down THROUGH $D$, and copy the circle radius $r$ onto that extension, so that you get a point $E$ on that extended line, but below the horizontal by amount $r$. Now the center $C$ must be equidistant from $A$ and $E$, so construct their perpendicular bisector; where it intersects that vertical line $\ell$ through $D$ is the center $C$.
This may seem pretty informal, but erecting perpendiculars, constructing bisectors, and copying lengths are all supported by pretty elementary theorems.
Picture:
Instructions using picture:
Erect the orange perpendicular to the baseline.
Draw a green radius in original circle
Draw parallel lines (faint grey) from one end of the green radius through $D$, and from the circle center through a point below $D$. Draw a line parallel to green radius at $D$, and truncate it by the two parallel lines to create a copy of the green segment at $D$.
Using $D$ and the copied green segment, construct the purple circle around $D$.
Construct the blue dot at the bottom of the picture: the itnersection of the orange line and the purple circle. There are two such; pick the one on the opposite side of the baseline from the original circle-center. Call that point $E$.
Construct the line $EA$ (red), and its perpendicular bisector (aqua).
The bisector meets the orange perpendicular at a point $C$; use this as a circle center, with radius $CD$. Use this to construct the thick black circle, which is tangent to both the circle at $A$ and to the line at $D$.
Idea: If we invert the situation at a circle around $D$, the given line remains unchanged (but $D$ is mapped to infinity) and the given circle turns into a circle (if we are lucky, it is the same circle again). But the desired circle is mapped to a line that is tangent to the (transformed) given circle and parallel to the given line. The two solutions to this can easily be constructed.
Construction: Let the line through $A$ perpendicular to the given line intersect the given circle in $P_1$ and $P_2$. Let the line $DP_k$ intersect the given circle a second time in $Q_k$. We want to make $Q_k$ the point where the circles touch. To this end, let $O$ be the intersection of $AQ_k$ with the line perpendicular to the given line through $D$. Then the circle around $O$ through $D$ solves the porblem.
Proof: The inversion at the circle around $D$ perpendicular to the given circle, leaves the given line and the given circle fixed. Also, it maps $P_k$ to $Q_k$ because it fixes the lines through $D$. Therefore, the parallel to the given lines through $P_k$ turns into a circle touching the given circle in $Q_k$ and passing through $D$. The centre $O$ of this circle must be both on $AQ_k$ and the perpendicular through $D$.
