# Thread: [SOLVED] Equation of Circle

1. ## [SOLVED] Equation of Circle

Im having trouble answering questions like this, any help would be much appreciated.

Find the centre and radius of the circle with equation

$x^2+y^2-6x-2y-6=0$

2. Originally Posted by edwards2779
Im having trouble answering questions like this, any help would be much appreciated.

Find the centre and radius of the circle with equation

$x^2+y^2-6x-2y-6=0$
x^2 - 6x + 9 + y^2 - 2y + 1 = 0 + 9 + 1 + 6

(x - 3)^2 + (y - 1)^2 = 16

I'll also give it to you in LaTeX. Just notice the constants I've marked in red.

$x^2 - 6x + 9 + y^2 - 2y + 1= 0 + 9 + 1 + 6$

$(x - 3)^2 + (y - 1)^2 = 16$

3. For
$x^2 +y^2 + 2hx +2fy +c =0$

Center is (-h,-f)

$
\sqrt{h^2 +f^2 -c}
$

EDIT: Too late Jan watch "6"

4. Originally Posted by edwards2779
Im having trouble answering questions like this, any help would be much appreciated.

Find the centre and radius of the circle with equation

$x^2+y^2-6x-2y-6=0$
If you are not familiar with the technique of completing the square, then you should review it.

In this example, we have

$x^2+y^2-6x-2y-6=0$

$\Rightarrow(x^2-6x+9)+(y^2-2y+1)=16$

$\Rightarrow(x-3)^2+(y-1)^2=4^2\text.$

Now can you see the center and radius?

5. oh i see now, so you basically just factorising. thanks for the help