What is x^2+y^2+4x-12y+15=0 ?

The solution to x^2+y^2+4x-12y+15=0 is Circle with (a,b)=(-2,6),r=5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^2+y^2+4x-12y+15=0

x^{2} + y^{2} + 4 x - 12 y + 15 = 0

(a, b) = (- 2, 6), r = 5

Solution steps

x^{2} + y^{2} + 4 x - 12 y + 15 = 0

Rewrite

x^{2} + y^{2} + 4 x - 12 y + 15 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 2, 6), r = 5

x^{2} + 2 x + y^{2} - 2 y + 1 = 0

x^{2} + 4 y + y^{2} - 12 = 0

(x - 5)^{(2)} + (y + 10)^{(2)} = 5^{(2)}

(x - 4)^{2} + (y - (- 3))^{2} = 5^{2}

(x - 1)^{2} + (y - 3)^{2} = 64

What is x^2+y^2+4x-12y+15=0 ?
The solution to x^2+y^2+4x-12y+15=0 is Circle with (a,b)=(-2,6),r=5