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

The solution to 9x^2-42x+9y^2+84y+65=0 is Circle with (a,b)=(7/3 ,-14/3),r=2sqrt(5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Rewrite

9 x^{2} - 42 x + 9 y^{2} + 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

{x - (- 3)}^{2} + {y - (- 4)}^{2} = 1

(x - 5)^{2} + (y + 2)^{2} = 7.5

x^{2} + (y - 7.5)^{2} = \frac{9}{49}

x^{2} + y^{2} - 2 x + 8 y = 0

x^{2} + y^{2} + 10 x - 4 y - 8 = 0

What is 9x^2-42x+9y^2+84y+65=0 ?
The solution to 9x^2-42x+9y^2+84y+65=0 is Circle with (a,b)=(7/3 ,-14/3),r=2sqrt(5)