How do you find the center and radius for $x^2+y^2+x-6y+9=0$ ?

Question

4925 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-28T04:46:17

center = $(-1/2,3)$

radius = $1/2$

We have the equation,

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

When a circle is in this form we must complete the square in order to put in a usable format. like,

$(x-a)^2+(y-b)^2=r^2$

where the centre is at $(a,b)$ and radius is r.

So rearrange,

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

And complete the square,

$(x+1/2)^2-1/4+(y-3)^2-9=-9$

Rearrange,

$(x+1/2)^2+(y-3)^2=1/4$

So from this equation, we know that our circle centre is at,

$(-1/2,3)$

and the radius of our circle is,

$sqrt(1/4)=1/2$

How do you find the center and radius for x^2+y^2+x-6y+9=0x^2+y^2+x-6y+9=0?