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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

Drawing

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.

N3buchadnezzar
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  • I'd say that the tangent in C is perpendicular to both $O_\beta$ and $O_\gamma$? – long tom Mar 21 '13 at 10:33
  • So $C$, $O_\gamma$ and $O_\beta$ are collinear? – long tom Mar 21 '13 at 10:36
  • i think the answer might be on http://math.stackexchange.com/questions/32386/finding-the-circles-passing-through-two-points-and-touching-a-circle?rq=1 (last part of accepted answer) – long tom Mar 21 '13 at 10:53
  • You're looking for a circle that is tangent to two given circles? – Sgernesto Mar 22 '13 at 12:52
  • 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 $C$.

    http://folk.ntnu.no/oistes/Diverse/sirkelmaple.pdf

     – N3buchadnezzar Mar 22 '13 at 13:05
  • "for most pairs $A,B$ there are two possible choices for $C$": for all pairs, surely? – TonyK Sep 15 '14 at 22:57

2 Answers2

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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.

Community
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user21820
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I will give you the general solution of the apollonius PPC which works fine for this case:

draw any circle $\Gamma$ that passes through $A$ and $B$ and meets $\gamma$ in two points $M$ and $N$. Let $H = MN \cap AB$ draw the tangents from $H$ to $\gamma$ and let $T$ and $S$ be the contact points of these tangent lines. The circles you want are the circles $\odot (TAB)$ and $\odot (SAB)$. The construction works because $H$ is by construction the radical center of $\gamma$, $\Gamma$ and the solutions and therefore lies in the common tangent line of the solutions and $\gamma$.

hellofriends
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