What is 4x^2-8x+4y^2-16y-124=0 ?

The solution to 4x^2-8x+4y^2-16y-124=0 is Circle with (a,b)=(1,2),r=6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4x^2-8x+4y^2-16y-124=0

4 x^{2} - 8 x + 4 y^{2} - 16 y - 124 = 0

(a, b) = (1, 2), r = 6

Solution steps

4 x^{2} - 8 x + 4 y^{2} - 16 y - 124 = 0

Rewrite

4 x^{2} - 8 x + 4 y^{2} - 16 y - 124 = 0

in the form of the standard circle equation

(x - 1)^{2} + (y - 2)^{2} = 6^{2}

Therefore the circle properties are:

(a, b) = (1, 2), r = 6

x^{2} + y^{2} - 14 x + 12 y + 59 = 0

x^{2} + y^{2} + 10 x + 4 y - 56 = 0

x^{2} + y^{2} - 5 x = 44 - 2 y

(x^{2} + 4 x) + (y^{2} + 4 y) = 9

x^{2} + y^{2} + 12 x + 2 y + \frac{1}{16} = 0

What is 4x^2-8x+4y^2-16y-124=0 ?
The solution to 4x^2-8x+4y^2-16y-124=0 is Circle with (a,b)=(1,2),r=6