Construct circle tangent to two lines Given two line segments $ab$ and $cd$, I want to draw a circle tangent to both line segments and passing through points $c$ and $b$.

Primitive operations available to me are:

• Draw a line between two points.
• Draw a perpendicular line passing through a point.
• Draw a line at a particular angle passing through a point.
• Construct a circle with a given line segment as the diameter.
• Construct a circle given 3 points on the boundary.
• Construct a circle given a center point and a point on the boundary.
• Find the midpoint of a line segment.
• Find center of an (already drawn) circle.

I can draw the circle passing through $c$ and $b$ and tangent to either $ab$ (or $cd$) by finding the intersection of a perpendicular bisector of $bc$ and a line from $b$ ($c$) perpendicular to $ab$ ($cd$).

However, I'm having trouble finding a circle passing through the points tangent to both line segments. Since I'm interested in actually drawing the such circles, obviously simpler constructions are better.

Update: Note that if there's a certain symmetry between the line segments, then the above construction works perfectly. Suppose we have a line segment $a'b$ instead of $ab$. Then we get the following: Find the centre of a circle passing through a known point and tangential to two known lines is a somewhat related question.

• As a rule of thumb, circles are determined by thee facts. You can give three points they pass through, you can give two points and tangency to a line, you can give tangency to two lines and a point, and so forth. You're requiring your circle satisfy four facts: pass through two points and be tangent to two lines. Your system is overdetermined. Hence TonyK's answer. – Neal May 16 '12 at 21:54
• @Neal The system isn't always over determined. For example, the two points/lines could both happen to be tangents on a unit circle. I drew several circles given similar constraints, but I guess the constraints just happened to be partially redundant. In particular, the perpendicular bisector of of the two points and the two perpendiculars to the two lines may all intersect at a point. – Joe May 16 '12 at 22:07
• No overdetermined system is always overdetermined. It's just almost always overdetermined :) – Neal May 18 '12 at 0:45

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.
• Well duh. Your "certain symmetry between the line segments" is my "$bx = cx$". – TonyK May 16 '12 at 22:29
• Then I don't understand your notation. $bx$ and $cx$ are not the same line. – Joe May 16 '12 at 22:30
• I mean $|bx| = |cx|$. – TonyK May 16 '12 at 22:31