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

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

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

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

(a, b) = (0, 0), r = 1

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (0, 0), r = 1

x^{2} + y^{2} - 4 y - 15 = 6

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

5 0^{2} = (x - 21.21)^{2} + (y - 21.21)^{2}

x^{2} + y^{2} - 4 x - 14 y + 50 = 0

9 x^{2} + 9 y^{2} + 72 y + 36 y + 155 = 0

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