What is (x-3)^2+(y^2-2^2)=5 ?

The solution to (x-3)^2+(y^2-2^2)=5 is Circle with (a,b)=(3,0),r=3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x-3)^2+(y^2-2^2)=5

(x - 3)^{2} + (y^{2} - 2^{2}) = 5

(a, b) = (3, 0), r = 3

Solution steps

(x - 3)^{2} + (y^{2} - 2^{2}) = 5

Rewrite

(x - 3)^{2} + (y^{2} - 2^{2}) = 5

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (3, 0), r = 3

16 = x^{2} + y^{2} - 4 x + 6 y + 13

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

x^{2} + y^{2} - 8 x + 10 y + 40 = 0

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

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

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