What is 16x^2+16y^2+32x+8y=47 ?

The solution to 16x^2+16y^2+32x+8y=47 is Circle with (a,b)=(-1,-1/4),r=2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

16x^2+16y^2+32x+8y=47

16 x^{2} + 16 y^{2} + 32 x + 8 y = 47

(a, b) = (- 1, - \frac{1}{4}), r = 2

Solution steps

16 x^{2} + 16 y^{2} + 32 x + 8 y = 47

Rewrite

16 x^{2} + 16 y^{2} + 32 x + 8 y = 47

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 1, - \frac{1}{4}), r = 2

What is 16x^2+16y^2+32x+8y=47 ?
The solution to 16x^2+16y^2+32x+8y=47 is Circle with (a,b)=(-1,-1/4),r=2