What is x^2+(y+1)^2=7 ?

The solution to x^2+(y+1)^2=7 is Circle with (a,b)=(0,-1),r=sqrt(7)

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x^2+(y+1)^2=7

x^{2} + (y + 1)^{2} = 7

(a, b) = (0, - 1), r = 7

Solution steps

x^{2} + (y + 1)^{2} = 7

Rewrite

x^{2} + (y + 1)^{2} = 7

in the form of the standard circle equation

(x - 0)^{2} + (y - (- 1))^{2} = (7)^{2}

Therefore the circle properties are:

(a, b) = (0, - 1), r = 7

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

x^{2} + (y + 1)^{2} = 9

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

(x + 6)^{2} + (y - 3)^{2} = 36

(x + 7)^{2} + (y - 7)^{2} = (5)^{2}

What is x^2+(y+1)^2=7 ?
The solution to x^2+(y+1)^2=7 is Circle with (a,b)=(0,-1),r=sqrt(7)