What is (x^2+3x)+(y^2-4y)=6 ?

The solution to (x^2+3x)+(y^2-4y)=6 is Circle with (a,b)=(-3/2 ,2),r= 7/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x^2+3x)+(y^2-4y)=6

(x^{2} + 3 x) + (y^{2} - 4 y) = 6

(a, b) = (- \frac{3}{2}, 2), r = \frac{7}{2}

Solution steps

x^{2} + 3 x + (y^{2} - 4 y) = 6

Rewrite

x^{2} + 3 x + (y^{2} - 4 y) = 6

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- \frac{3}{2}, 2), r = \frac{7}{2}

2 x^{2} + 2 y^{2} - 2 x + 4 y = 45

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

x^{2} + y^{2} + 8 x + 8 y = - 16

3 x^{2} + 6 y + 3 y^{2} + 5 x - 15 = 0

x^{2} + y^{2} - 2 x + 4 y = 11

What is (x^2+3x)+(y^2-4y)=6 ?
The solution to (x^2+3x)+(y^2-4y)=6 is Circle with (a,b)=(-3/2 ,2),r= 7/2