What is 4x^2+4y^2+36y+5=0 ?

The solution to 4x^2+4y^2+36y+5=0 is Circle with (a,b)=(0,-9/2),r=sqrt(19)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

4 x^{2} + 4 y^{2} + 36 y + 5 = 0

(a, b) = (0, - \frac{9}{2}), r = 19

Solution steps

4 x^{2} + 4 y^{2} + 36 y + 5 = 0

Rewrite

4 x^{2} + 4 y^{2} + 36 y + 5 = 0

in the form of the standard circle equation

(x - 0)^{2} + (y - (- \frac{9}{2}))^{2} = (19)^{2}

Therefore the circle properties are:

(a, b) = (0, - \frac{9}{2}), r = 19

x^{2} + \frac{y ^{2}}{64} = 1

\frac{x ^{2}}{9} + \frac{y ^{2}}{9} = 1

x = y^{2} - 1

\frac{( x - 3 ) ^{2}}{81} - \frac{( y + 5 ) ^{2}}{144} = 1

x = \frac{1}{4} y^{2}

What is 4x^2+4y^2+36y+5=0 ?
The solution to 4x^2+4y^2+36y+5=0 is Circle with (a,b)=(0,-9/2),r=sqrt(19)