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

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

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

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

(a, b) = (- 4, 3), r = 3

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 4, 3), r = 3

x^{2} + (y + 4)^{2} = \frac{1}{9}

x^{2} + (y - 6)^{2} = 9

\frac{( x - 2 ) ^{2}}{16} + \frac{( y + 5 ) ^{2}}{16} = 1

x^{2} - 2 x + 1 + y^{2} + y + \frac{1}{2} = \frac{11}{4}

x^{2} + 2 x + y^{2} - 2 y - 45 = 0

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