I need to find the circle that will be tangent to a line at a given point and a circle. The diagram below hopefully makes it clearer.


The known data are:
points P2,P3,P4,P5
Circle C2's radius r
Angle C
The distance from P3 to P2, a

I need to find P1 or b (which makes it easy enough to work out P1). There are actually two solutions to this problem. Circle C1 could also be tangent on the far side of C2 (c=b-r).

As some background info I am using this to find the medial axis of a polygon. By stepping P3 along each line/arc of the polygon I can generate an approximation of the medial axis. I have already worked out the solution for two lines.


The set of points that are equidistant from the circle $C_2$ and the line $P_4P_5$ is a parabola. Translate the line $P_4P_5$ (in the opposite direction than the $P_2$) by the radius $r$ of the circle (i.e. $|P_4'P_4| = r$, $|P_5'P_5| = r$ and $P_4P_5\ \|\ P_4'P_5'$) to get a new line $P_4'P_5'$ and observe that your solution will be equidistant from the line $P_4'P_5'$ and the point $P_2$ (if before the distance was $b$, then now it is $b+r$). Precisely, the $P_2$ will be the focus and $P_4'P_5'$ will be the directrix. To find the point $P_1$ of parabola just construct the perpendicular bisector of $P_3'P_2$, it should intersect $P_3'P_3$ exactly in $P_1$.

Edit: Added some pictures (the parabola below is just an approximation, not a real parabola).



  • $\begingroup$ Thanks. Translating the line makes everything easy. Translating it the other way gives the second solution. $\endgroup$ – Les Newell May 24 '12 at 14:43
  • $\begingroup$ @LesNewell Yes, exactly ;-) Just that this direction of translation is easier to imagine (and so was my choice). $\endgroup$ – dtldarek May 24 '12 at 14:55
  • $\begingroup$ Thank you for your detailed help. It would be nice to try to calculate the parabola directly but things get complicated as the circle is actually just an arc and it is joined to other arcs/lines. $\endgroup$ – Les Newell May 24 '12 at 16:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.