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

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

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

(x - 1)^{2} + y^{2} \leq 1

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

Solution steps

(x - 1)^{2} + y^{2} \leq 1

Rewrite

(x - 1)^{2} + y^{2} \leq 1

in the form of the standard circle equation

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

Therefore the circle properties are:

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

(x - 2)^{2} + (y + 7)^{2} = 49

x^{2} + y^{2} + 8 x + 12 y = 0

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

(x - 2)^{2} + (y + 2)^{2} = 20

(x - 2)^{2} + (y + 2)^{2} = 25

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