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

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

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(y-1)^2+x^2<= 4

(y - 1)^{2} + x^{2} \leq 4

(a, b) = (0, 1), r = 2

Solution steps

(y - 1)^{2} + x^{2} \leq 4

Rewrite

(y - 1)^{2} + x^{2} \leq 4

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (0, 1), r = 2

(x - 8)^{2} + (y - 6)^{2} = 1 0^{2}

(5 (x + 4))^{2} + 25 (y - 2)^{2} = 625

x^{2} + y^{2} - 14 x + 12 y + 76 = 0

x^{2} + y^{2} + 6 x + 24 y + 152 = 0

x^{2} + y^{2} - 3 x + 8 y = 20

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