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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

x^{2} + y^{2} + 6 x + 8 x = - 16

x^{2} + y^{2} - 10 x - 6 y + 25 = 0

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

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

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

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