What is the center x^2+y^2-4x-6y-23=0 ?

The center x^2+y^2-4x-6y-23=0 is (2,3)

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-6y-23=0

center

x^{2} + y^{2} - 4 x - 6 y - 23 = 0

(2, 3)

Solution steps

x^{2} + y^{2} - 4 x - 6 y - 23 = 0

Rewrite

x^{2} + y^{2} - 4 x - 6 y - 23 = 0

in the form of the standard circle equation

(x - 2)^{2} + (y - 3)^{2} = 6^{2}

Therefore the circle properties are:

(a, b) = (2, 3), r = 6

And the center is:

(2, 3)

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

(x - 2)^{2} + (y - 3)^{2} = 49

(x - 5)^{2} + (y + 7)^{2} = 4

3 x^{2} + 3 y^{2} - 12 x - 2 = 0

(x - (- 5))^{2} + (y - (- 6))^{2} = \frac{2}{3}