What is (x+2)^2+(y+1)^2=2 ?

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

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(x+2)^2+(y+1)^2=2

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

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

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

x^{2} + 4 x + y^{2} + 10 y = - 20

x^{2} + y^{2} + 10 x + 6 y - 47 = 0

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

(x - 1 + 23)^{2} + (y - 3)^{2} = 20

x^{2} + y^{2} + 4 x + 2 y - 27 = 0

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