What is 4(x-5)^2+4y^2=20 ?

The solution to 4(x-5)^2+4y^2=20 is Circle with (a,b)=(5,0),r=sqrt(5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4(x-5)^2+4y^2=20

4 (x - 5)^{2} + 4 y^{2} = 20

(a, b) = (5, 0), r = 5

Solution steps

4 (x - 5)^{2} + 4 y^{2} = 20

Rewrite

4 (x - 5)^{2} + 4 y^{2} = 20

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (5, 0), r = 5

x^{2} + y^{2} - 3 x + 5 y - 1 = 0

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

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

(x - 1)^{2} + (y - 3)^{2} = 13

x^{2} + y^{2} + 8 x - 12 y - 8 = 0

What is 4(x-5)^2+4y^2=20 ?
The solution to 4(x-5)^2+4y^2=20 is Circle with (a,b)=(5,0),r=sqrt(5)