Apologies, but I didnt like MathManiacs answer. To say the least its confusingly written, and fails to carry a construction through to the end, and makes assumptions about what is provided. Some vital details seem to be brushed over. I attempted to follow his construction and failed; I dont know why. I couldnt see what direction he was going and the instructions themselves dont make a whole lot of sense to me when you actually try to carry it out. Perhaps its my failure but I thought Id give a different perspective, filling some gaps. I of course dont mean any disrespect to the community but I didnt feel as though this 3 year old question was ever truly answered. I only came across this question in an Ask Jeeves search because I particularly love geometry.
Here is my own construction. Hopefully you can follow along with the animated GIF image. This isn't a proof; it's just a method of construction.
We have an arbitrary line m, and an arbitrary line segment AB defining the center, A, and the radius AB of a circle. No circle is drawn there because circle A(B) is of any size and you only have a rusty compass, which is defined for you in the upper-right corner. We wish to find the points of intersection on line m of where circle A(B) intersects the line.
- Segment AB can be extended to line m, intersecting it at C.
- Naturally for this construction to work line AB cannot be parallel with line m. See the notes toward the bottom for dealing with this case.
- Draw a circle A(r) centered at point A. This circle will intersect the extended line AB at some point D.
- Draw a completely arbitrary line n, passing through point A but not coincident with AB, intersecting the constant circle A(r) at point E.
- Points D and E define a new line, DE.
- Construct two new lines, parallel to DE, passing through B and C, intersecting line n at points F and G, respectively.
- Points D and F define a new line, DF.
- Construct a new line parallel to DF, but passing through G. This line intersects the extended line AB at point H.
- Draw a line parallel to line m but passing through H. This line intersects the constant circle A(r) at two points X1 and X2.
- Lines AX1 and AX2 intersect the line m at two points I1 and I2. These are your desired intersections.
If X1 and X2 dont exist then neither do the intersections I1 and I2, for obvious reasons.
I have no citation for this construction. It is based on the notion of shrinking down the scale, so that the circle A(B) becomes A(r), and line m becomes a new line that can be directly intersected with A(r) (producing the X intersections). The two I points - the intersections of interest - are projections of the two X points, away from the point A and onto the line m again at appropriate scale. Circle A(r) is to circle A(B) what the horizontal line passing through H is to the line m. The bulk of the construction is scaling line m down in proportion.
Creating a Parallel Line
To draw a parallel line using a straightedge and a single circle, the Wikipedia has an article entitled Poncelet-Steiner Theorem which contains an animated GIF depicting the construction. It requires 5 intermediary lines to be drawn for each parallel line to be made, and 1 circle (of any size) for each line from which a parallel is to be made.
The Case of Parallel AB
In the case that line segment AB is parallel to line m, a separate construction could be implemented to rotate AB into a new non-parallel position. You then restart this line-circle intersection construction using a new line segment AB'.
That said, the construction for this case is actually easier than just explained. You dont need to rotate AB at all. I wont show you the modification, but you can swap the roles of lines m and n in the original construction and adapt the construction appropriately. This can be done fairly readily and with no additional cost.
I absolutely love geometric constructions and in particular restricted constructions. Send me more. The aforementioned Poncelet-Steiner theorem, and of course the Mohr-Mascheroni Theorem, are fascinating restricted constructions in their own right.
Best Answer
Here it goes,
It must be a necessary condition that two given points lie on Same side of given line
Now, Let the two given points be $P$ and $Q$, let the given line be $L$
Case-1-When segment $PQ$ is not parallel to line $L$
Join points $P$ and $Q$ and extend it to cut the line $L$ at point $S$ Measure distance $PS$ using compass,Here in my case,$P$ is nearer to point $S$ than $Q$,
Now mark a point on extended line on Other side Of segment $PS$ with a distance equal to that of $PS$,let this new point be $R$,
Draw perpendicular Bisector of Segment $RQ$,Mark point $G$ where it intersects the segment $RQ$
Taking $G$ as centre and $GR$ as radius,Draw a circle $A$
Draw a perpendicular $X$ Passing through point $S$,Mark point $F$ where it intersects the circle $A$
Measure Length $SF$ using compass or ruler,Mark Point $Z$ on given line $L$ along initial side of $S$ with length $SF$ measured
The Point $Z$ is the point of tangency Of required Circle, Now Draw Any perpendicular Bisectors of segments out of Segment $PQ$,Segment $QZ$ and Segment $PZ$
Mark the intersection of these any two perpendicular Bisectors as point $K$,it is the Centre of Required Circle
Finally draw a circle Using Found centre and three Points, I will be uploading a Geogebra Image soon ,I don't have enough Time to type..
Line in orange color is original line given,$PQZ$ is the required Circle,$Q$ and $P$ are given points
Case-2- When $PQ$ is parallel to line $L$, Draw Perpendicular bisector line $J$ of segment $PQ$ ,Mark it's intersection point with Line $L$ as $O$, this point $O$ is point of tangency, Repeat last steps of case -1 to find required Circle
Hope this helps
EDIT
I have added an image for More Understanding, Now let's understand what we have done geometrically in first case
The basic property used is the secant tangent theoram of circle,see here
Now as the basic idea is clear,we needed to find the geometric mean of $SP$ and $SQ$ equivalent to geometric mean of $SR$ and $SQ$ which is precisely the length of $SZ$
Now, To find geometric mean, We have a nice property in right angle triangle,see here
In order to use this,We needed to create a right angled triangle,and we know that in a circle, Any traingle with one point on Circle and base as Diameter make One nice right triangle,So we made circle with diameter $RQ$ ,which we created using point $G$ or midpoint as centre,
Later we dropped a perpendicular from $S$ to new circle to use Geometric mean theoram ,as mentioned ,We found the geometric mean to be precisely the length of $SF$ ,hence we marked point $Z$ on line $L$ such that $SF=SZ$
Hence finding the point of tangency,Other than this, It is now simple to draw a circle knowing $3$ Points on it