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

x^{2} + y^{2} = 200

X^{2} + Y^{2} - 5 X + 4 Y = 46

(x + 4)^{2} + (y - 4)^{2} = 8

x^{(2)} + y^{(2)} + 2 x - 7 = 0

x^{2} + y^{2} = 2 5^{2}

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