How do you find the center and radius for #x^2 + y^2 + 3x + 6y +9=0#?

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Answer 1 · 2016-08-05T09:09:17

#C(-3/2,-3)#; #r=3/2#

#x^2+y^2+ax+by+c=0#

The center is obtained by:

#C(-a/2,-b/2)#

and the radius by:

#r=1/2sqrt(a^2+b^2-4c)#

Then let's apply the to the case:

#x^2+y^2+3x+6y+9=0#

and have

#C(-3/2,-3)#

#r=1/2sqrt(9+cancel36-cancel36)=3/2#