What is the center x^2+y^2-4x+10y+20=0 ?

The center x^2+y^2-4x+10y+20=0 is (2,-5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

center x^2+y^2-4x+10y+20=0

center

x^{2} + y^{2} - 4 x + 10 y + 20 = 0

(2, - 5)

Solution steps

x^{2} + y^{2} - 4 x + 10 y + 20 = 0

Rewrite

x^{2} + y^{2} - 4 x + 10 y + 20 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (2, - 5), r = 3

And the center is:

(2, - 5)

x^{2} + y^{2} - 7 x = 36 - 4 y

4 x^{2} + 4 y^{2} = 9

x^{2} + y^{2} - 10 x + 10 y + 49 = 0

x^{2} - 4 x + y^{2} + 6 y + 12 = 0

x^{2} + y^{2} - 10 x + 20 y + 61 = 0

What is the center x^2+y^2-4x+10y+20=0 ?
The center x^2+y^2-4x+10y+20=0 is (2,-5)