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Let $C_{1}$ be a circle of unit radius. Let A and B be two points inside $C_{1}$. Now I want to construct another circle $C_{2}$ such that A and B lie on $C_{2}$ and $C_{2}$ is orthogonal to $C_{1}$ at their point of intersection(I want $C_{2}$ in such a way that it intersects $C_{1}$). I tried and failed to find a way to construct such a $C_{2}$.

Any help is appreciated.

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  • $\begingroup$ need not be..for example construct $C_{1}$ and $C_{2}$ such that they are orthogonal. Now select 2 points on $C_{2}$ whose bisector need not goes through center of $C_{1}$ $\endgroup$ – Phani Raj Jan 29 '13 at 17:16
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    $\begingroup$ My roundabout and inefficient (nay, stupid) way of doing this: apply the fractional-linear map that turns the unit disk into the upper half-plane. Now the problem is easily solved. Of course you have to transform the circle you’ve found back to the original setting. $\endgroup$ – Lubin Jan 29 '13 at 17:33
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    $\begingroup$ Take the inverse of $A$ with respect to the circle $C_1$ to obtain the third point $C$. Now construct the circle containing $A,B,C$. $\endgroup$ – Sigur Jan 29 '13 at 17:34
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    $\begingroup$ @Sigur, very elegant! $\endgroup$ – Lubin Jan 29 '13 at 17:36
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If I understand your question correctly, it is answered in Construction 2.1 of this paper: http://comp.uark.edu/~strauss/papers/hypcomp.pdf

Edit upon request: Given two points $A$ and $B$ in the interior of a circle $C_1$, invert $A$ through $C_1$ to $A'$. Now construct the circle through $A$, $B$, and $A'$. Sigur's answer contains a picture of this construction.

The paper discusses general compass-and-straightedge constructions in hyperbolic geometry, where, in particular, geodesics can be modeled as circles perpendicular to the unit circle in $\mathbb{C}$.

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    $\begingroup$ Nice paper! Thanks. $\endgroup$ – Sigur Jan 29 '13 at 17:38
  • $\begingroup$ While the linked pdf may solve the question, it is not very helpful as an answer since the links may go stale. I know it's twelve pages long, but could you at least vaguely summarize it here? $\endgroup$ – Tobias Kienzler Mar 20 '13 at 7:57
  • $\begingroup$ @TobiasKienzler As requested. $\endgroup$ – Neal Mar 20 '13 at 13:20
  • $\begingroup$ Thanks, now the answer deserves the checkmark $\endgroup$ – Tobias Kienzler Mar 20 '13 at 13:50
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Take the inverse of $A$ with respect to the circle $C_1$ (see here) to obtain the third point $C$. Now construct the circle containing $A,B,C$.

enter image description here

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