What is 4x^{(2)}+4y^{(2)}-8y-5=0 ?

The solution to 4x^{(2)}+4y^{(2)}-8y-5=0 is Circle with (a,b)=(0,1),r= 3/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4x^{(2)}+4y^{(2)}-8y-5=0

4 x^{(2)} + 4 y^{(2)} - 8 y - 5 = 0

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

Solution steps

4 x^{2} + 4 y^{2} - 8 y - 5 = 0

Rewrite

4 x^{2} + 4 y^{2} - 8 y - 5 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

4 y^{2} + 24 y + 4 x^{2} - 12 + 9 = 0

(x + 8)^{2} + (y - 6)^{2} = 100

2 x^{2} + 2 y^{2} + 6 x - 8 y - 3 = 0

3 0^{2} = a^{2} + b^{2}

x^{2} + y^{2} + 6 x - 8 y + 5 = 0

What is 4x^{(2)}+4y^{(2)}-8y-5=0 ?
The solution to 4x^{(2)}+4y^{(2)}-8y-5=0 is Circle with (a,b)=(0,1),r= 3/2