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