1. ## Algerbil

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?

(x-4)^2 +4^2 + (x-2)^2 - 2^2 +11 = 0
At first, this is not a circle, it's just a parabola.

May be you mean $(x-4)^2+(y-2)^2=-23,$ which is clearly, an imaginary circle.

3. Originally Posted by Krizalid
At first, this is not a circle, it's just a parabola.

May be you mean $(x-4)^2+(y-2)^2=-23,$ which is clearly, an imaginary circle.
Here's what I started with:

x^2 - 8x + y^2 + 4y + 11 = 0

I completed teh square for the first couple of expressions and got to where I was in my first post.

Here's what I started with:

x^2 - 8x + y^2 + 4y + 11 = 0

I completed teh square for the first couple of expressions and got to where I was in my first post.
that's incorrect then

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

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

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

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

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