What is (x+3)^2+(y-1)^2=64 ?

The solution to (x+3)^2+(y-1)^2=64 is Circle with (a,b)=(-3,1),r=8

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(x+3)^2+(y-1)^2=64

(x + 3)^{2} + (y - 1)^{2} = 64

(a, b) = (- 3, 1), r = 8

Solution steps

(x + 3)^{2} + (y - 1)^{2} = 64

Rewrite

(x + 3)^{2} + (y - 1)^{2} = 64

in the form of the standard circle equation

(x - (- 3))^{2} + (y - 1)^{2} = 8^{2}

Therefore the circle properties are:

(a, b) = (- 3, 1), r = 8

x^{2} + y^{2} + 2 x - 6 y + 5 = 0

x^{2} + y^{2} - 6 x + 4 y - 20 = 0

2 (x + 6)^{2} + 2 (y + 9)^{2} = 50

2 x^{2} + 2 y^{2} + 16 y + 14 = 0

(x - 19)^{2} + (y + 18)^{2} = 289

What is (x+3)^2+(y-1)^2=64 ?
The solution to (x+3)^2+(y-1)^2=64 is Circle with (a,b)=(-3,1),r=8