What is the center x^2+y^2-14x-12y+59=0 ?

The center x^2+y^2-14x-12y+59=0 is (7,6)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

center x^2+y^2-14x-12y+59=0

center

(7, 6)

Solution steps

x^{2} + y^{2} - 14 x - 12 y + 59 = 0

Rewrite

x^{2} + y^{2} - 14 x - 12 y + 59 = 0

in the form of the standard circle equation

(x - 7)^{2} + (y - 6)^{2} = (26)^{2}

Therefore the circle properties are:

(a, b) = (7, 6), r = 26

And the center is:

(7, 6)

What is the center x^2+y^2-14x-12y+59=0 ?
The center x^2+y^2-14x-12y+59=0 is (7,6)