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

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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#