What is the center x^2+y^2= 1/4 ?

The center x^2+y^2= 1/4 is (0,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

center x^2+y^2= 1/4

center

(0, 0)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - 0)^{2} + (y - 0)^{2} = (\frac{1}{2})^{2}

Therefore the circle properties are:

(a, b) = (0, 0), r = \frac{1}{2}

And the center is:

(0, 0)