What is 4x^2+4y^2-4x+2y-1=0 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4x^2+4y^2-4x+2y-1=0

4 x^{2} + 4 y^{2} - 4 x + 2 y - 1 = 0

(a, b) = (\frac{1}{2}, - \frac{1}{4}), r = \frac{3}{4}

Solution steps

4 x^{2} + 4 y^{2} - 4 x + 2 y - 1 = 0

Rewrite

4 x^{2} + 4 y^{2} - 4 x + 2 y - 1 = 0

in the form of the standard circle equation

(x - \frac{1}{2})^{2} + (y - (- \frac{1}{4}))^{2} = (\frac{3}{4})^{2}

Therefore the circle properties are:

(a, b) = (\frac{1}{2}, - \frac{1}{4}), r = \frac{3}{4}

25 x^{2} + 16 y^{2} = 400

\frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1

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

x^{2} + y^{2} = 121

\frac{x ^{2}}{9} + \frac{y ^{2}}{25} = 1

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