What is x^2+y^2-4x+14y+37=0 ?

The solution to x^2+y^2-4x+14y+37=0 is Circle with (a,b)=(2,-7),r=4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

x^2+y^2-4x+14y+37=0

x^{2} + y^{2} - 4 x + 14 y + 37 = 0

(a, b) = (2, - 7), r = 4

Solution steps

x^{2} + y^{2} - 4 x + 14 y + 37 = 0

Rewrite

x^{2} + y^{2} - 4 x + 14 y + 37 = 0

in the form of the standard circle equation

(x - 2)^{2} + (y - (- 7))^{2} = 4^{2}

Therefore the circle properties are:

(a, b) = (2, - 7), r = 4

What is x^2+y^2-4x+14y+37=0 ?
The solution to x^2+y^2-4x+14y+37=0 is Circle with (a,b)=(2,-7),r=4