What is (x-40)^2+(y-0)^2=0.1 ?

The solution to (x-40)^2+(y-0)^2=0.1 is Circle with (a,b)=(40,0),r=0.31622…

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(x-40)^2+(y-0)^2=0.1

(x - 40)^{2} + (y - 0)^{2} = 0.1

(a, b) = (40, 0), r = 0.31622 \dots

Solution steps

(x - 40)^{2} + (y - 0)^{2} = 0.1

Rewrite

(x - 40)^{2} + (y - 0)^{2} = 0.1

in the form of the standard circle equation

(x - 40)^{2} + (y - 0)^{2} = 0.31622 \dots^{2}

Therefore the circle properties are:

(a, b) = (40, 0), r = 0.31622 \dots

3 x^{2} - 2 x + 3 + 3 y^{2} + 8 y + 48 = \frac{172}{3}

(x - 1)^{2} + (y + 5)^{2} < 4

(X + 4)^{2} + (Y - 1)^{2} = 148

(x + 2)^{2} + (y + 3)^{2} = 11

x^{2} + y^{2} + 4 x - 4 y = 1

What is (x-40)^2+(y-0)^2=0.1 ?
The solution to (x-40)^2+(y-0)^2=0.1 is Circle with (a,b)=(40,0),r=0.31622…