Write the equation of the circle in the form $(x-h)^{2}+(y-k)^{2}=r^{2}$ $x^{2}+y^{2}+6 x+2 y+3=0$.
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Step 1 ：The general form of the equation of a circle is \(x^{2}+y^{2}+2gx+2fy+c=0\). Comparing this with the given equation, we can see that \(g=3\), \(f=1\) and \(c=3\).

Step 2 ：The center of the circle is given by \((-g, -f)\) and the radius is given by \(\sqrt{g^{2}+f^{2}-c}\).

Step 3 ：We can substitute these values into the equations to find the center and radius of the circle.

Step 4 ：Substituting the values we get \(h = -3\), \(k = -1\), and \(r = 2.6457513110645907\).

Step 5 ：The equation of the circle in the form \((x-h)^{2}+(y-k)^{2}=r^{2}\) is \((x+3)^{2}+(y+1)^{2}=7\).

Step 6 ：Final Answer: The equation of the circle in the form \((x-h)^{2}+(y-k)^{2}=r^{2}\) is \(\boxed{7}\).