What is (x-3)^2+y^2<4 ?

The solution to (x-3)^2+y^2<4 is Circle with (a,b)=(3,0),r=2

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(x-3)^2+y^2<4

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

(a, b) = (3, 0), r = 2

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - 3)^{2} + (y - 0)^{2} = 2^{2}

Therefore the circle properties are:

(a, b) = (3, 0), r = 2

3 x^{2} + 3 y^{2} + 36 x - 24 y + 96 = 0

x^{2} + y^{2} - 20 x - 14 y + 100 = 0

(x - 4)^{2} + (Y (x) + 2)^{2} = 4

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

\frac{x ^{2}}{2} + \frac{y ^{2}}{2} + 112. 6^{2} = 180. 3^{2}

What is (x-3)^2+y^2<4 ?
The solution to (x-3)^2+y^2<4 is Circle with (a,b)=(3,0),r=2