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

The solution to 4x^2+12x+4y^2+16-11=0 is Circle with (a,b)=(-3/2 ,0),r=1

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+16-11=0

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

(a, b) = (- \frac{3}{2}, 0), r = 1

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- \frac{3}{2}))^{2} + (y - 0)^{2} = 1^{2}

Therefore the circle properties are:

(a, b) = (- \frac{3}{2}, 0), r = 1

x^{2} = 50 - y^{2}

215 = a^{2} + b^{2}

(x)^{2} + (y - 4)^{2} = 73

x^{2} + y^{2} - x - y - \frac{1}{2} = 0

(x - 5)^{2} + (y + 4)^{2} = 100

What is 4x^2+12x+4y^2+16-11=0 ?
The solution to 4x^2+12x+4y^2+16-11=0 is Circle with (a,b)=(-3/2 ,0),r=1