If AB is a common tangent to two circles, prove that the circle on AB as a diameter cuts each of the circles orthogonally.
Source: Challenge and Thrills in Pre College Mathematics.
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$\begingroup$ Are $A$ and $B$ the tangent points? $\endgroup$– A.Γ.Sep 22, 2015 at 8:48
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1 Answer
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Proving Orthognality at points A and B is trivial as it follows from the definition of a tangent (and the angle that it forms with the radius of the circle drawn through the point of contact).
What remains to be proven is that the other two points of intersection A' and B' of the circle with Diameter AB with the original two circles are such that OA' and OB' are tangents where O is the centre of the third circle.
Let P be the centre of one of the circles whose tangents are drawn OA and OA' It is easy to prove that triangle OAP and OA'P are congruent by S-S-S test. Hence, angle OAP=angle OA'P=90 degrees. Thus, OA' is also a tangent, which completes the proof.
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$\begingroup$ Ah I missed naming A and B on the common tangent.. Kindly assume it to be implied $\endgroup$ Sep 22, 2015 at 8:43
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1$\begingroup$ Why the downvote? What constructive purpose does it serve if a downvote is not accompanied by a clarification/improvement suggestion in the comments section? $\endgroup$ Sep 22, 2015 at 8:45
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$\begingroup$ Looks good. Perhaps it is a bit better to consider just one tangent circle to avoid discussion whether the circles are on the same or the opposite sides of the tangent. $\endgroup$– A.Γ.Sep 22, 2015 at 8:56
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$\begingroup$ I wanted to remove the other common tangents, but unfortunately, geogebra app allowed me to either add all of them or none. (I am a first time user of that app, so there may be a way of removing them that I am not aware of). $\endgroup$ Sep 22, 2015 at 9:01
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$\begingroup$ Also, there is no need to avoid discussion of whether the circles are on the same or the opposite sides of the tangent. The textual answer written above is generic and applies to all cases except when points A and B are not distinct (which is out of the scope for this question). For the sake of completeness, the textual answer should be supplemented with a diagram for the other case too, but for representational purposes, I have left that for the OP to do. $\endgroup$ Sep 22, 2015 at 9:03