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

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

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(x-3)^2+(y+2)^2=3

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

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

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

(x - 2)^{2} + (y - 3)^{2} = 25

\frac{2 x ^{2} + 2 y ^{2} - 3 x + 5 y - 3}{2} = 0

(x + 12)^{2} + (y - 4)^{2} - 30 = 0

6 x^{2} + 6 y^{2} - 24 x + 36 y = 72

x^{2} + y^{2} - 2 (x + y) > 0

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