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While browsing around about problems similar to the problem of Apollonius, I have found references to constructions of all types of circles. For example, not only is it possible to construct a circle tangent to three given circles, but one can construct a circle through any three points, tangent to any three lines, passing through two given points and tangent to a line or circle, passing through a given point and tangent to two given lines or circles, etc. Pretty much any combination of criteria regarding points, lines, and circles, with repetition.

Can we have one of each? I haven't found any resource saying whether it's possible or not to construct a circle through a given point, tangent to a given line, and tangent to a given circle. Is such a construction possible? Thanks.

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In general, you need three "conditions" to be able to draw a circle (consistent with the fact that the Cartesian equation for the circle has three undetermined parameters). The wonder behind things like the "nine point circle" is that circles set to satisfy only three conditions happen to satisfy a bunch of other conditions, too. –  J. M. Apr 28 '11 at 0:32
    
Three conditions determine a circle, but to be sure that they are constructible, one would have to constrain the type of condition. –  Phira Apr 28 '11 at 0:41
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If the line separates the point from the circle, or if the circle separates the point from the line, then: no. –  Blue Apr 28 '11 at 1:10
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3 Answers 3

up vote 4 down vote accepted

It is possible, provided that the point and at least part of the circle are on the same side of the line and the circle does not separate the point from the line. If the point and the circle are on opposite sides of the line, or if the point is inside the circle and the line wholly outside, then it is impossible (as it is with three nested circles in the original version).

Wikipedia describes this as Special case 6 of Apollonius' problem with up to 4 solutions

example of four tangents

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Thanks for the image Henry, and the link to that wikipedia article seems to be just what I was looking for. –  yunone Apr 28 '11 at 1:41
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If the circle and the line have a point of intersection, you can send the point of intersection to infinity with an inversion (with respect to a circle which takes this point as center) and you have reduced it to a problem with two tangent lines and a point (which I presume that you know).

For the case without intersection, I would have to reflect further, but in any case, it is equivalent by inversion to one point and two circles if you have already accepted the cases with repetition.

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Thanks for your input, user9325. –  yunone Apr 28 '11 at 1:42
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i will use inversion to solve the problem. let the given point is p, circle c, and line l. call a circle centered at p and unit radius $\omega.$ call the inverses of $c^\prime$ and $l^\prime.$ find the common tangents to (there could be as many as four) to $c^\prime$ and $l^\prime$ invert each of the tangents back on $\omega$ that should give you the required circles.

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