What is a^2+b^2-1<= 2a ?

The solution to a^2+b^2-1<= 2a is Circle with (a,b)=(1,0),r=sqrt(2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

a^2+b^2-1<= 2a

a^{2} + b^{2} - 1 \leq 2 a

(a, b) = (1, 0), r = 2

Solution steps

a^{2} + b^{2} - 1 \leq 2 a

Rewrite

a^{2} + b^{2} - 1 \leq 2 a

in the form of the standard circle equation

(a - 1)^{2} + (b - 0)^{2} = (2)^{2}

Therefore the circle properties are:

(a, b) = (1, 0), r = 2

2 x^{(2)} + 2 y^{(2)} - 3 x + 4 y - 1 = 0

4 x^{2} + 4 y^{2} - 24 x + 8 y + 39 = 0

(x - 4)^{2} + (y - 5)^{2} = 64

(x)^{2} + (y)^{2} - 6 x + 10 y + 34 = 72

a - b + a^{2} + b^{2} = 0

What is a^2+b^2-1<= 2a ?
The solution to a^2+b^2-1<= 2a is Circle with (a,b)=(1,0),r=sqrt(2)