What is the center 9x^2+9y^2+42x+84y+65=0 ?

The center 9x^2+9y^2+42x+84y+65=0 is (-7/3 ,-14/3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

center 9x^2+9y^2+42x+84y+65=0

center

(- \frac{7}{3}, - \frac{14}{3})

Solution steps

9 x^{2} + 9 y^{2} + 42 x + 84 y + 65 = 0

Rewrite

9 x^{2} + 9 y^{2} + 42 x + 84 y + 65 = 0

in the form of the standard circle equation

(x - (- \frac{7}{3}))^{2} + (y - (- \frac{14}{3}))^{2} = (25)^{2}

Therefore the circle properties are:

(a, b) = (- \frac{7}{3}, - \frac{14}{3}), r = 25

And the center is:

(- \frac{7}{3}, - \frac{14}{3})

What is the center 9x^2+9y^2+42x+84y+65=0 ?
The center 9x^2+9y^2+42x+84y+65=0 is (-7/3 ,-14/3)