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Re: circle inversion is orthogonal
Posted:
Jan 22, 2002 5:52 AM
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"rdack" <rdacker@pacbell.net> escribiÃÂó en el mensaje
news://644f6688.0201211656.6ee47fe0@posting.google.com...
> 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'
R: radius of circle C'
r: radius of circle C
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 CaÃÂñestro
A CoruÃÂña (EspaÃÂña)
ignacio.larrosa@eresmas.net
ICQ 94732648
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