What is 4x^2+4y^2-32y+39=0 ?

The solution to 4x^2+4y^2-32y+39=0 is Circle with (a,b)=(0,4),r= 5/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

4x^2+4y^2-32y+39=0

4 x^{2} + 4 y^{2} - 32 y + 39 = 0

(a, b) = (0, 4), r = \frac{5}{2}

Solution steps

4 x^{2} + 4 y^{2} - 32 y + 39 = 0

Rewrite

4 x^{2} + 4 y^{2} - 32 y + 39 = 0

in the form of the standard circle equation

(x - 0)^{2} + (y - 4)^{2} = (\frac{5}{2})^{2}

Therefore the circle properties are:

(a, b) = (0, 4), r = \frac{5}{2}

What is 4x^2+4y^2-32y+39=0 ?
The solution to 4x^2+4y^2-32y+39=0 is Circle with (a,b)=(0,4),r= 5/2