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