What is 4x^2+12x+4y^2+16y+11=0 ?

The solution to 4x^2+12x+4y^2+16y+11=0 is Circle with (a,b)=(-3/2 ,-2),r=sqrt(7/2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4x^2+12x+4y^2+16y+11=0

4 x^{2} + 12 x + 4 y^{2} + 16 y + 11 = 0

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

Solution steps

4 x^{2} + 12 x + 4 y^{2} + 16 y + 11 = 0

Rewrite

4 x^{2} + 12 x + 4 y^{2} + 16 y + 11 = 0

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}

(x + 5)^{2} + (y + 6)^{2} = 12

x^{2} + y^{2} - x - y - 16 = 0

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

x^{2} + y^{2} - 4 x - 6 y - 36 = 0

(x + 9)^{2} + (y - 8)^{2} = 81

What is 4x^2+12x+4y^2+16y+11=0 ?
The solution to 4x^2+12x+4y^2+16y+11=0 is Circle with (a,b)=(-3/2 ,-2),r=sqrt(7/2)