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

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

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

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

(a, b) = (0, 0), r = 22

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - 0)^{2} + (y - 0)^{2} = (22)^{2}

Therefore the circle properties are:

(a, b) = (0, 0), r = 22

(x - 1)^{2} + (y - 3)^{2} = 121

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

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

(x + 3)^{2} + (y - 1)^{2} = 6

9 x^{2} + 9 y^{2} + (- 12) x + 6 y + 1 = 0

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