I have to simplify this to get the equation of a circle:

(x-4)^2 +4^2 + (x-2)^2 - 2^2 +11 = 0

I get:

(x-4)^2 + (x-2)^2 +16 + 4 + 11 = 0

(x-4)^2 + (x-2)^2 = -31

....where am I going wrong?

Results 1 to 4 of 4

- Nov 6th 2007, 11:12 AM #1

- Nov 6th 2007, 11:16 AM #2

- Nov 6th 2007, 11:22 AM #3

- Nov 6th 2007, 12:12 PM #4
that's incorrect then

$\displaystyle x^2 - 8x + y^2 + 4y + 11 = 0$

$\displaystyle \Rightarrow x^2 - 8x + y^2 + 4y = -11$

$\displaystyle \Rightarrow x^2 - 8x + (-4)^2 + y^2 + 4y + 2^2 = -11 + (-4)^2 + 2^2$

$\displaystyle \Rightarrow (x - 4)^2 + (y + 2)^2 = 3^2$

which is a circle centered at (4,-2) with radius 3