Here is a construction of mine that takes a square root but uses some other tricks to reduce the number of steps in the overall construction. There probably is a better construction but this is my best so far. This version is closer to the graphic you have in your question and combines everything into one graphic.
I will take it as already shown that
$$\frac 1{\sqrt{a}}+\frac 1{\sqrt{b}}=\frac 1{\sqrt{c}}$$
In the diagram, your two given circles are drawn in red ($a'$ with center $A$ and radius $AP=a$) and blue ($b'$ with center $B$ and radius $BP=b$). Point $P$ is the intersection of segment $\overline{AB}$ with the circles.
Construct semicircle $d$ between points $A$ and $B$, and point $D$ on semicircle $d$ so that line $\overleftrightarrow{DP}$ is perpendicular to $\overline{AB}$. We then have
$$DP=\sqrt{ab}$$
as can be shown by considering the tree right triangles formed by points $A,B,P,$ and $D$.
In the next stage, construct ray $h$ to bisect angle $BPD$ and point $F$ at its intersection with segment $\overline{BD}$, and drop a perpendicular from point $F$ to segment $\overline{BP}$ intersecting at point $G$. Length $FG=PG$ is the reciprocal of the sum of the reciprocals of $BP$ and $DP$. By construction and the formula at the start of this answer, we have
$$FG=PG=\sqrt{bc}$$
In the next stage, construct quarter-circle $f$ with center $P$ and radius $PG$, point $H$ at the intersection of quarter-circle $f$ with line $\overleftrightarrow{DP}$, and point $J$ on segment $\overline{PH}$ such that segment $\overline{GJ}$ is parallel to segment $\overline{BH}$. By similar triangles we then have $\frac{GJ}{BH}=\frac{PJ}{PH}$, which means that
$$PJ=c$$
which is the desired radius of the third circle tangent to the line.
In the last stage, construct a semicircle with center $P$ and radius $PJ$, point $N$ the intersection of that semicircle with segment $AP$, point $O$ the intersection of that semicircle with segment $BP$, the circular arc $o$ with center $A$ and radius $AO$, the circular arc $n$ with center $B$ and radius $BN$, and point $C$ the intersection of arcs $n$ and $o$. Draw the circle $c'$ with center $C$ and radius length $PJ$, and you now have that third circle!
Draw the line tangent to any two of those circles, which you say you can do, and you are all done. I hope you can see in my diagram that the three circles do indeed have the same tangent line $t$.
Best Answer
It is possible, provided that the point and at least part of the circle are on the same side of the line and the circle does not separate the point from the line. If the point and the circle are on opposite sides of the line, or if the point is inside the circle and the line wholly outside, then it is impossible (as it is with three nested circles in the original version).
Wikipedia describes this as Special case 6 of Apollonius' problem with up to 4 solutions