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

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

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2y=y^2+(x+4)^2

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

(a, b) = (- 4, 1), r = 1

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 4, 1), r = 1

x^{2} + (y - 7)^{2} = 9

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

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

(x - 3)^{2} + (y - 5)^{2} = 9

y^{2} = 6 + x - x^{2}

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