What is x^2+y^2-4^2-12=0 ?

The solution to x^2+y^2-4^2-12=0 is Circle with (a,b)=(0,0),r=2sqrt(7)

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x^2+y^2-4^2-12=0

x^{2} + y^{2} - 4^{2} - 12 = 0

(a, b) = (0, 0), r = 27

Solution steps

x^{2} + y^{2} - 4^{2} - 12 = 0

Rewrite

x^{2} + y^{2} - 4^{2} - 12 = 0

in the form of the standard circle equation

(x - 0)^{2} + (y - 0)^{2} = (27)^{2}

Therefore the circle properties are:

(a, b) = (0, 0), r = 27

a^{2} + b^{2} - 1 \leq 2 a

2 x^{(2)} + 2 y^{(2)} - 3 x + 4 y - 1 = 0

4 x^{2} + 4 y^{2} - 24 x + 8 y + 39 = 0

(x - 4)^{2} + (y - 5)^{2} = 64

(x)^{2} + (y)^{2} - 6 x + 10 y + 34 = 72

What is x^2+y^2-4^2-12=0 ?
The solution to x^2+y^2-4^2-12=0 is Circle with (a,b)=(0,0),r=2sqrt(7)