What is (x+3)^2+(y-4)^2=16 ?

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

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(x+3)^2+(y-4)^2=16

(x + 3)^{2} + (y - 4)^{2} = 16

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

Solution steps

(x + 3)^{2} + (y - 4)^{2} = 16

Rewrite

(x + 3)^{2} + (y - 4)^{2} = 16

in the form of the standard circle equation

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

Therefore the circle properties are:

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

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