What is (x-2)^2+(y-3)^2=1 ?

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

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

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

(a, b) = (2, 3), r = 1

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (2, 3), r = 1

x^{2} + (y - 8)^{2} = 81

(x - 4)^{2} + (y - 4)^{2} = 9

(x - 4)^{2} + (y - 4)^{2} = 5

4 x^{2} + 4 y^{2} - 4 x + 12 y - 71 = 0

x, x^{2} + Y (x)^{2} = 64

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