What is (x-(-2))^2+(y-1)^2=4 ?

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

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(x-(-2))^2+(y-1)^2=4

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

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

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

x^{2} + y^{2} \leq 64

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

x^{2} + y^{2} + 2 x - 2 y + 1 = 0

y^{2} + x^{2} = 64

y^{2} + x^{2} = 72

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