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Topic: circle inversion is orthogonal
Replies: 10   Last Post: Jan 24, 2002 5:43 PM

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 ilarrosa@linuxfan.com Posts: 258 Registered: 12/6/04
Re: circle inversion is orthogonal
Posted: Jan 22, 2002 5:52 AM

"rdack" <rdacker@pacbell.net> escribiÃÂÃÂ³ en el mensaje
> i want to prove:
> a circle passing through both points of a circle inversion pair
> will be orthogonal to the circle of inversion.
>
> i see references to it. even a proof which draws a tangent to the
> circle with the inversion pair from the center of the circle of
> inversion. it then says it is obvious by similar triangles. what
> similar triangles? i mean, i see the two triangles that must be
> similar, but not how they are similar.
>
> if P and P' are the inversion pair of circle C, and A is the tangent
> point on C from the center of inversion circle K, then triangle OAP
> must be similar to OAP'. but why? the only fact i have is that line OA
> is perpendicular to line from A to center of C by construction. must
> be another bit of info there, somewhere.
>
> Once we have similar triangles, OA must be radius of K, so the two
> circles are orthogonal.
>
> i'm sure it's obvious. just not to me.

K: centre of inversion circle C'
O: centre of circle C, passing through P and P'

d: distance KO

Then, if P (external) an P' (internal) are inverse with respect to C',

KP*KP'=R^2 (by definition of inversion)

(d+r)(d-r)=R^2 ==> d^2 - r^2 = R^2 ==>

d^2 = R^2 + r^2 ==> angle KAO = 90 deg.

The radius KA and OA are perpendicular, and so the circles are orthogonal

--
Saludos,

Ignacio Larrosa CaÃÂÃÂ±estro
A CoruÃÂÃÂ±a (EspaÃÂÃÂ±a)
ignacio.larrosa@eresmas.net
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