What is 1^2=(x^2)/4+(y^2)/4 ?

The solution to 1^2=(x^2)/4+(y^2)/4 is Circle with (a,b)=(0,0),r=2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

1^2=(x^2)/4+(y^2)/4

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

(a, b) = (0, 0), r = 2

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (0, 0), r = 2

(x + 3)^{2} + (y - 4)^{2} = 1

x^{2} + y^{2} + 6 x + 6 y = 0

x^{2} - 4 x + y^{2} + 14 y = 11

x^{2} + y^{2} - 14 x + 18 y - 12 = 0

(x - 1)^{2} + y^{2} \leq 1

What is 1^2=(x^2)/4+(y^2)/4 ?
The solution to 1^2=(x^2)/4+(y^2)/4 is Circle with (a,b)=(0,0),r=2