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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

(x - \frac{3}{5})^{2} + (y + \frac{7}{8})^{2} = \frac{9}{4}

x^{2} + y^{2} - 4 x + 5 y - 5 = 0

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

(x - 7)^{2} + (y - 3)^{2} = 4

5 x^{2} + 5 y^{2} + 100 x + 50 y + 220 = 0

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