What is (x-3)^2+(y+13)^2=9 ?

The solution to (x-3)^2+(y+13)^2=9 is Circle with (a,b)=(3,-13),r=3

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(x-3)^2+(y+13)^2=9

(x - 3)^{2} + (y + 13)^{2} = 9

(a, b) = (3, - 13), r = 3

Solution steps

(x - 3)^{2} + (y + 13)^{2} = 9

Rewrite

(x - 3)^{2} + (y + 13)^{2} = 9

in the form of the standard circle equation

(x - 3)^{2} + (y - (- 13))^{2} = 3^{2}

Therefore the circle properties are:

(a, b) = (3, - 13), r = 3

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

x^{2} + y^{2} + 9 x + 10 = 0

4 y^{2} - 72 y = 316 - 4 x^{2} - 24 x

(x - \frac{1}{2})^{2} + (y + \frac{1}{4})^{2} = 12

x^{(2)} + y^{(2)} + 4 x - 10 y = - 16

What is (x-3)^2+(y+13)^2=9 ?
The solution to (x-3)^2+(y+13)^2=9 is Circle with (a,b)=(3,-13),r=3