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MathGroup Archive 1998

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Re: new user help

  • To: mathgroup at smc.vnet.net
  • Subject: [mg13471] Re: [mg13296] new user help
  • From: "Fred Simons" <wsgbfs at win.tue.nl>
  • Date: Sun, 26 Jul 1998 02:33:33 -0400
  • Sender: owner-wri-mathgroup at wolfram.com

John,

If you see beforehand (and everybody should!) that the midpoint is  (-2,
3), you can apply the substitutions x->-2 + h, y->3+k, expand  the
result and then substitute back h->x+2, k->y+3.

But there exists a much nicer and more general way in doing it. Put

exp1 = x^2 + y^2 + 4 x - 6 y + 4

exp2 = (x-h)^2 + (y-k)^2 - r^2

These two expressions have to be equal for all x and y, so we apply

SolveAlways[ exp1 == exp2, {x, y} ]

and arrive at

{{r -> -3, h -> -2, k -> 3},  {r -> 3, h -> -2, k -> 3}}

These two solutions have the same values for h, k and r^2, so the 
result is

exp2 /. %[[1]]

-9 + (2 + x)^2 + (-3 + y)^2

Hope this helps,

Fred Simons
Eindhoven University of Technology


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