What is (x-3)^2+(y+7)^2=16 ?

The solution to (x-3)^2+(y+7)^2=16 is Circle with (a,b)=(3,-7),r=4

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(x-3)^2+(y+7)^2=16

(x - 3)^{2} + (y + 7)^{2} = 16

(a, b) = (3, - 7), r = 4

Solution steps

(x - 3)^{2} + (y + 7)^{2} = 16

Rewrite

(x - 3)^{2} + (y + 7)^{2} = 16

in the form of the standard circle equation

(x - 3)^{2} + (y - (- 7))^{2} = 4^{2}

Therefore the circle properties are:

(a, b) = (3, - 7), r = 4

(x - \frac{5}{2})^{2} + (y - \frac{1}{2})^{2} = \frac{169}{4}

(x - 13)^{2} + (y + 13)^{2} = 16

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

x^{2} + y^{2} - 18 x + 16 y + 120 = 0

x^{2} + y^{2} + 8 x - 12 y + 47 = 0

What is (x-3)^2+(y+7)^2=16 ?
The solution to (x-3)^2+(y+7)^2=16 is Circle with (a,b)=(3,-7),r=4