The full details of this problem is given as follows

Construct a circle $\gamma$ with center $O_\gamma$ , and place two points $A$ and $B$ inside $\gamma$. That does not lie on the edge of the circle. Explain the construction of a point $C$, such that the circle $ABC =\beta$, is internally tangential to $\gamma$.

Now $ABC$ means a circle that passes through the points $A$,$B$ and $C$. I have made a drawing, but I am unable to mathematicaly construct the point $C$. I already know that for most pairs $A$,$B$ there are two possible choices for $C$. Eg $C_1$ and $C_2$. See the following figure


Can anyone show me or help me in finding the placement of $C$, given $A$ and $B$? The figure is only but a sketch, but I know that the centre of the circle obviously has to lie on the perpendicular bisector of A and B, after that I am clueless.

  • $\begingroup$ I'd say that the tangent in C is perpendicular to both $O_\beta$ and $O_\gamma$? $\endgroup$ – long tom Mar 21 '13 at 10:33
  • $\begingroup$ So $C$, $O_\gamma$ and $O_\beta$ are collinear? $\endgroup$ – long tom Mar 21 '13 at 10:36
  • $\begingroup$ i think the answer might be on math.stackexchange.com/questions/32386/… (last part of accepted answer) $\endgroup$ – long tom Mar 21 '13 at 10:53
  • $\begingroup$ You're looking for a circle that is tangent to two given circles? $\endgroup$ – Sgernesto Mar 22 '13 at 12:52
  • $\begingroup$ Given two points A and B inside a circle $\gamma$. Find a point $C$ that lies on $\gamma$, such that the circle that passes through $A$,$B$,$C$ only touches $\gamma$ at $C$. folk.ntnu.no/oistes/Diverse/sirkelmaple.pdf $\endgroup$ – N3buchadnezzar Mar 22 '13 at 13:05

I gave an answer here, which applies to the general case of finding a circle tangent to three circles, which may have up to 8 solutions. This particular case is a degenerate one where two of the circles are points, and has only 2 solutions. It can be solved by the last step of the general solution I gave, which is to invert at one of the two points, and the desired circle will become a line that must be tangent to a circle and pass through a point, which is easy to construct. Inversion can be easily constructed using compass and straightedge also, and the original desired solutions can likewise be obtained by undoing the inversion.


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