What is 16=(x-2)^2+(y+3)^2 ?

The solution to 16=(x-2)^2+(y+3)^2 is Circle with (a,b)=(2,-3),r=4

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16=(x-2)^2+(y+3)^2

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

(a, b) = (2, - 3), r = 4

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - 2)^{2} + (y - (- 3))^{2} = 4^{2}

Therefore the circle properties are:

(a, b) = (2, - 3), r = 4

2 x^{2} + 2 y^{2} + 8 x - 12 y - 26 = 0

x^{2} + y^{2} + 12 x - 14 y + 36 = 0

x^{2} + y^{2} - 2 x - 6 y + \frac{1}{4} = 0

x^{2} + 16 x + y^{2} + 18 y = - 120

x^{2} + y^{2} - 12 x - 2 y + 12 = 0

What is 16=(x-2)^2+(y+3)^2 ?
The solution to 16=(x-2)^2+(y+3)^2 is Circle with (a,b)=(2,-3),r=4