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

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Answer 1 · 2017-02-22T20:05:33

Centre: #(0, -9/2)#

Radius: #sqrt 19#

You complete two squares :)

For #4x^2+4y^2+36y+5=0#, tactically i find it easier to remove the coefficient on the quadratic, so we go at this instead:

#4 (x^2+y^2+9y+5/4)=0#

Or:
# x^2+color(red)(y^2+9y)+5/4=0#

Completing the square in #y#:
# x^2+color(red)(((y+ 9/2)^2 - 81/4 ))+5/4=0#

#implies x^2+(y+ 9/2)^2 - 19=0#

Or:

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

#implies x^2+(y+ 9/2)^2 = sqrt( 19)^2#