What is (x-10)^2+(y+1)^2-64=0 ?

The solution to (x-10)^2+(y+1)^2-64=0 is Circle with (a,b)=(10,-1),r=8

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(x-10)^2+(y+1)^2-64=0

(x - 10)^{2} + (y + 1)^{2} - 64 = 0

(a, b) = (10, - 1), r = 8

Solution steps

(x - 10)^{2} + (y + 1)^{2} - 64 = 0

Rewrite

(x - 10)^{2} + (y + 1)^{2} - 64 = 0

in the form of the standard circle equation

(x - 10)^{2} + (y - (- 1))^{2} = 8^{2}

Therefore the circle properties are:

(a, b) = (10, - 1), r = 8

- 3 y^{2} = 3 x^{2} - 2 x - 1

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

2 x^{2} + 2 y^{2} - 28 x + 12 y + 114 = 0

(x - 4)^{2} + (y + 11)^{2} = 25

x^{2} + y^{2} - 3 = 1

What is (x-10)^2+(y+1)^2-64=0 ?
The solution to (x-10)^2+(y+1)^2-64=0 is Circle with (a,b)=(10,-1),r=8