My Math Forum Geometry: Circle Inversion and Power of P

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February 15th, 2010, 02:33 PM   #1
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Geometry: Circle Inversion and Power of P

Alright.. I'm going over circle inversion and power of P, for some reason something isn't clicking.

The question is: (pretend the numerical values are the little sub values)

Show that the product of PP1 and PP2 (or PQ1 and PQ2) can be expressed as PO^(2) - r^(2).

Where P is the Power of P.

I got the image off some website, so it's not the exact one that I have. Pretend that where O is, it's actually P.. and pretend that there is a point in the center of the circle called O.
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 February 16th, 2010, 05:30 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,624 Thanks: 2077 I've seen that website; it does make sense, but perhaps the following will help. You acknowledge that PP1/PQ1 = PQ2/PP2, which is true since triangles PQ1P2 and PP1Q2 are similar. That still applies when Q1 and Q2 are the same point, Q, and so P1.P2 = PQ² = PO² - r² by Pythagoras, where r is the radius of the circle.

 Tags circle, geometry, inversion, power

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