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=3

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 = 3

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} = 3^{2}

Therefore the circle properties are:

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

4 x^{2} + 4 y^{2} = 1

4 x^{2} + 4 y^{2} + 6 x - 8 y - 4 = 0

x^{2} + y^{2} - 10 x - 12 y - 3 = 0

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

x^{2} + y^{2} = \frac{1}{4}

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=3