What is x^2+y^2-2x-4y+4<= 0 ?

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

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x^2+y^2-2x-4y+4<= 0

x^{2} + y^{2} - 2 x - 4 y + 4 \leq 0

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

Solution steps

x^{2} + y^{2} - 2 x - 4 y + 4 \leq 0

Rewrite

x^{2} + y^{2} - 2 x - 4 y + 4 \leq 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

x^{2} + y^{2} + 6 x - 18 y + 25 = 0

5 = ((x + 1)^{2} + (y - (- 5))^{2})^{2}

x^{(2)} - 6 x + y^{(2)} + 4 y = 3

x^{2} + y^{2} - 6 x + 12 y + 20 = 0

x^{2} + 8 x + y^{2} - 2 y + 17 = 4 x^{2} - 24 x + 4 y^{2} - 16 y + 52

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