euclidean-geometrygeometric-constructiongeometry
Using a ruler and a compass how can construct a line through a point and tangent to a circle. What I don't want is to eyeball the line by trying to line-up the ruler over the circle. Best if I could construct the point of intersection first and then draw the line.
PS. I know how to do it mathematically, I just don't know the steps for geometry, given A, C and the circle to find D.
Update
Based on answers here is the constructors. Thanks for the quick responses.
If $x$ is the point where your two lines meet, then there is no such circle unless $bx = cx$. Which is why you're having trouble.
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
Draw a Thales circle over the segment $AC$, it will intersect the desired $D$, because $AD\perp DC$: