What is x^2+y^2-4x-2y-3=0 ?

The solution to x^2+y^2-4x-2y-3=0 is Circle with (a,b)=(2,1),r=2sqrt(2)

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x^2+y^2-4x-2y-3=0

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

(a, b) = (2, 1), r = 22

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - 2)^{2} + (y - 1)^{2} = (22)^{2}

Therefore the circle properties are:

(a, b) = (2, 1), r = 22

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

x^{2} + y^{2} - \frac{1}{4} x + \frac{1}{4} y = \frac{7}{32}

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

3 x^{2} + 3 y^{2} - 4 x + y - 12 = 0

x^{2} + y^{2} \leq 10

What is x^2+y^2-4x-2y-3=0 ?
The solution to x^2+y^2-4x-2y-3=0 is Circle with (a,b)=(2,1),r=2sqrt(2)