What is (x-0.04)^2+(y-0.069)^2=0.21^2 ?

The solution to (x-0.04)^2+(y-0.069)^2=0.21^2 is Circle with (a,b)=(0.04,0.069),r=0.21

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(x-0.04)^2+(y-0.069)^2=0.21^2

(x - 0.04)^{2} + (y - 0.069)^{2} = 0.2 1^{2}

(a, b) = (0.04, 0.069), r = 0.21

Solution steps

(x - 0.04)^{2} + (y - 0.069)^{2} = 0.2 1^{2}

Rewrite

(x - 0.04)^{2} + (y - 0.069)^{2} = 0.2 1^{2}

in the form of the standard circle equation

(x - 0.04)^{2} + (y - 0.069)^{2} = 0.2 1^{2}

Therefore the circle properties are:

(a, b) = (0.04, 0.069), r = 0.21

x (x - 6) + y (y + 4) = 3

x^{2} + y^{2} - 18 x + 6 y + 81 = 0

x^{2} + y^{2} - 4 x + 12 y - 8 = 0

(x + 3)^{2} + (y - 3)^{2} = 9

x^{2} + y^{2} + 18 x - 6 y + 54 = 0

What is (x-0.04)^2+(y-0.069)^2=0.21^2 ?
The solution to (x-0.04)^2+(y-0.069)^2=0.21^2 is Circle with (a,b)=(0.04,0.069),r=0.21