What is the center x^2+y^2<= 9 ?

The center x^2+y^2<= 9 is (0,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

center x^2+y^2<= 9

center

x^{2} + y^{2} \leq 9

(0, 0)

Solution steps

x^{2} + y^{2} \leq 9

Rewrite

x^{2} + y^{2} \leq 9

in the form of the standard circle equation

(x - 0)^{2} + (y - 0)^{2} = 3^{2}

Therefore the circle properties are:

(a, b) = (0, 0), r = 3

And the center is:

(0, 0)

x^{2} + 10 x + y^{2} - 16 - 11 = 0

x^{2} + y^{2} + 10 x - 8 y + 25 = 0

4 x^{2} + 4 y^{2} = 144

(x + 1)^{2} + y^{2} = 246

(x - 4)^{2} + (y + 1^{2})^{2} = 9