What is (x-1)^2+(y+6)^2<= 4 ?

The solution to (x-1)^2+(y+6)^2<= 4 is Circle with (a,b)=(1,-6),r=2

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(x-1)^2+(y+6)^2<= 4

(x - 1)^{2} + (y + 6)^{2} \leq 4

(a, b) = (1, - 6), r = 2

Solution steps

(x - 1)^{2} + (y + 6)^{2} \leq 4

Rewrite

(x - 1)^{2} + (y + 6)^{2} \leq 4

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (1, - 6), r = 2

(x - 10)^{2} + (y + 1)^{2} - 64 = 0

- 3 y^{2} = 3 x^{2} - 2 x - 1

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

2 x^{2} + 2 y^{2} - 28 x + 12 y + 114 = 0

(x - 4)^{2} + (y + 11)^{2} = 25

What is (x-1)^2+(y+6)^2<= 4 ?
The solution to (x-1)^2+(y+6)^2<= 4 is Circle with (a,b)=(1,-6),r=2