What is (x+4)^2+(y-2)^2=16 ?

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

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(x+4)^2+(y-2)^2=16

(x + 4)^{2} + (y - 2)^{2} = 16

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

Solution steps

(x + 4)^{2} + (y - 2)^{2} = 16

Rewrite

(x + 4)^{2} + (y - 2)^{2} = 16

in the form of the standard circle equation

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

Therefore the circle properties are:

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

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

x^{2} + y^{2} - 10 x + 6 y - 15 = 0

4 (x - 5)^{2} + 4 y^{2} = 64

3 x^{2} - 6 x + 3 y^{2} - 18 y - 78 = 0

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

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