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sujoy
09-10-2005, 03:33 AM
4 e.q., are given as:
Px^2+Qx+R = 0............................[1]
P=a^2+b^2.......................[2]
Q= 2(ab+ac)......................[3]
R= a^2+c^2......................[4]
now, can we draw "2" circles with P & R, how would they look like :? :!: :?:
regards
sujoy

Denis
09-10-2005, 03:58 AM
They will look like this: o, O :roll:

Gene
09-10-2005, 05:06 AM
I am getting suspicious of every thing you type. The usual circle is
x²+y²=r²
using the x axis and y axis.
You seem to be using the a axis and b axis and the a axis and c axis. That would mean either two 2D graphs or one 3D graph.

Denis
09-10-2005, 08:58 AM
Ahhhso, sqrt(P) and sqrt(R) would be hypotenuses, no?

Gene
09-10-2005, 01:00 PM
Well, I would prefer radii but I'm still confused about the axes.
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Gene

Denis
09-10-2005, 03:20 PM
Px^2+Qx+R = 0....................[1]
P=a^2+b^2.......................[2]
Q= 2(ab+ac)......................[3]
R= a^2+c^2......................[4]
now, can we draw "2" circles with P & R, how would they look like :?

Well, using quadratic on [1]:

x = [-Q +- sqrt(Q^2 - 4PR)] / 2P

Substituting [2][3][4] leads to (unless I goofed):

x = {-a(b + c) +- sqrt[-(a^4 - 2a^2bc + b^2c^2)]} / (a^2 + b^2)
x = {-a(b + c) +- sqrt[-(a^2 - bc)^2]} / (a^2 + b^2)

But I'm not sure how to handle the discriminant...
Well, even if I did, whadda **** would I do with results anyway!?

sujoy
09-11-2005, 08:43 AM
there was a problem which came like this since any part of a circle might represent real nos.....[HM.Hardy]..... I thought if a physical significance can be founf My concepts would be clearer
regards