If this can easily be derived from an already answered question please be so kind to point me there. I tried but couldn't find an equivalent problem.
I have one circle $D$ with the origin as center fully defined by its radius $r_D$. I have a given point $P$ which is outside $D$.
Now I want to construct a circle $H$ with known radius $r_H > r_D$, such that $P$ is on the circle and $D$ is enclosed inside $H$ and $H$ is touching $D$.
Where is the circle center of $H$/how do I calculate the point where $D=H$?
There are two solutions (though I just need one).
Notice that $H$ must touch (i.e. not intersect) $D$. This means it's center is located on a circle around the center of $D = (0,0)$ with the radius $r_H - r_D$. Simultaniously it must be on a circle with radius $r_D$ with center $P$. Now your problem is reduced to finding the intersection points of two circles.
Say: $(c_x,c_y)$ is the center of $H$ and wlog $P = (P_x,0)$. You have the equations: $$ c_x^2+c_y^2 = (r_D - r_H)^2 \text{ and } (c_x - P_x)^2+c_y^2 = r_H^2 $$ The rest should be simple, i guess.