Coordinate geometry -HELP PLEASE
- Thread starter DBashev
- Start date
Lets do the general solution and see if that is enough:
Suppose we have an equation
a x2 + a y2 + bx + c y + d = 0
where a is not zero and b, c, and d are real valued. Well if a is not zero, we can divide through by a and get
x2 + y2 + bx + c y + d = 0
where I have just relabeled the other constants, i.e. replaced b/a with b, etc.
We now need to complete the squares for the x and y parts. We'll do x and y can be done the same way:
x2 + bx = x2 + 2 (b/2) x + (b/2)2 - (b/2)2 = (x + b/2)2 - (b/2)2
Doing the same for y we can now write our original equation as
x2 + y2 + bx + c y + d = (x + b/2)2 - (b/2)2 + (y + c/2)2 - (c/2)2 + d = 0
or
(x + b/2)2 + (y + c/2)2 = (b/2)2 + (c/2)2 - d
Letting
r = \(\displaystyle \frac{1}{2}\sqrt{b^2 + c^2 - 4 d}\)
we finally have the equation of a circle [IF r is real, i.e. \(\displaystyle b^2 + c^2 \ge 4 d\)],
(x - x0)2 + (y - y0)2 = r2
with center at (x0, y0) and radius r where
x0 = -b/2
y0 = -c/2
As an aside, if \(\displaystyle b^2 + c^2 \lt 4 d\) we would still have a circle but x and y would be complex numbers with real parts x0 and y0 and imaginary parts satisfying a circle of radius [EDIT: magnitude of] r centered (0,0) for the imaginary part.
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- Feb 4, 2004
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There are loads of online resources showing the method. Try this listing. Once you have studied at least two lessons from the list, please attempt the exercise. If you get stuck, you can then reply with a clear listing of your efforts so far.
Thank you!
