What is x^2+y^2-22x+6y+129=0 ?

The solution to x^2+y^2-22x+6y+129=0 is Circle with (a,b)=(11,-3),r=1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^2+y^2-22x+6y+129=0

x^{2} + y^{2} - 22 x + 6 y + 129 = 0

(a, b) = (11, - 3), r = 1

Solution steps

x^{2} + y^{2} - 22 x + 6 y + 129 = 0

Rewrite

x^{2} + y^{2} - 22 x + 6 y + 129 = 0

in the form of the standard circle equation

(x - 11)^{2} + (y - (- 3))^{2} = 1^{2}

Therefore the circle properties are:

(a, b) = (11, - 3), r = 1

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

2 x^{2} + 2 y^{2} + 44 x + 28 y = 52

x^{2} + (y - 8)^{2} = 5

(x - 6)^{2} + (y - 4)^{2} = 5^{2}

25.314 4^{2} = (27.2 - x)^{2} + (27.2 - y)^{2}

What is x^2+y^2-22x+6y+129=0 ?
The solution to x^2+y^2-22x+6y+129=0 is Circle with (a,b)=(11,-3),r=1