What is (x-7)^2+(y+2)^2=16 ?

The solution to (x-7)^2+(y+2)^2=16 is Circle with (a,b)=(7,-2),r=4

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(x-7)^2+(y+2)^2=16

(x - 7)^{2} + (y + 2)^{2} = 16

(a, b) = (7, - 2), r = 4

Solution steps

(x - 7)^{2} + (y + 2)^{2} = 16

Rewrite

(x - 7)^{2} + (y + 2)^{2} = 16

in the form of the standard circle equation

(x - 7)^{2} + (y - (- 2))^{2} = 4^{2}

Therefore the circle properties are:

(a, b) = (7, - 2), r = 4

x^{2} + y^{2} - 2 x + 8 y + 13 = 0

(x + 7)^{2} + (y + 8)^{2} = \frac{4}{9}

2 y + 13 + x^{2} = 8 x - y^{2}

2 x^{2} + 2 y^{2} + 8 x - 6 y + 4 = 0

(x - 5)^{2} + (y + 2)^{2} = 9

What is (x-7)^2+(y+2)^2=16 ?
The solution to (x-7)^2+(y+2)^2=16 is Circle with (a,b)=(7,-2),r=4