How do you find the center and radius of the circle given #4x^2+4y^2+36y+5=0#?

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Answer 1 · 2017-01-31T07:10:13

The centre of the circle is #(0,-9/2)# and the radius #=sqrt19#

We complete the squares

#4x^2+4y^2+36y+5=0#

#4x^2+4(y^2+9y)=-5#

#4x^2+4(y^2+9y+81/4)=-5+81#

#4x^2+4(y+9/2)^2=76#

Dividing by #4#

#x^2+(y+9/2)^2=76/4=19#

So the centre of the circle is #(0,-9/2)# and the radius #=sqrt19#
graph{4x^2+4y^2+36y+5=0 [-13.72, 14.76, -13.18, 1.07]}