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:

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

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

How do you find the center and radius for 4x^2+4y^2+36y+5=04x^2+4y^2+36y+5=0?