What is (x+2)^2+(y-2)^2=16 ?

The solution to (x+2)^2+(y-2)^2=16 is Circle with (a,b)=(-2,2),r=4

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(x+2)^2+(y-2)^2=16

(x + 2)^{2} + (y - 2)^{2} = 16

(a, b) = (- 2, 2), r = 4

Solution steps

(x + 2)^{2} + (y - 2)^{2} = 16

Rewrite

(x + 2)^{2} + (y - 2)^{2} = 16

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 2, 2), r = 4

(x + 5)^{2} + (y - 2)^{2} = 61

(x + 2)^{2} + (y + \frac{1}{2})^{2} = \frac{17}{4}

(2 x^{2} - 12 x) + (2 y^{2} + 8 y) = 24

2 x^{2} + 2 y^{2} + 4 x - 8 y - 8 = 0

x^{(2)} + y^{(2)} = \frac{( 9 )}{( 4 )}

What is (x+2)^2+(y-2)^2=16 ?
The solution to (x+2)^2+(y-2)^2=16 is Circle with (a,b)=(-2,2),r=4