Question 942828
Smallest? There's only one solution.
Use the x and y intercepts. 
They form a right triangle with a circle inscribed in that right triangle.
x-intercept : {{{y=0}}}
{{{x-2sqrt(3)=0}}}
{{{x=2sqrt(3)}}}
y-intercept : {{{x=0}}}
{{{sqrt(3)y-2sqrt(3)=0}}}
{{{y=2}}}
So you have a right triangle triangle with one leg, {{{2sqrt(3)}}}, and another leg, {{{2}}}. 
Find the hypotenuse using the Pythagorean theorem,
{{{(2sqrt(3))^2+2^2=H^2}}}
{{{12+4=H^2}}}
{{{H^2=16}}}
{{{H=4}}}
For an inscribed circle in a right triangle,
{{{R=(A+B-H)/2}}}
{{{R=(2sqrt(3)+2-4)/2}}}
{{{R=sqrt(3)+1-2}}}
{{{R=sqrt(3)-1}}}
So the equation of the circle is,
{{{(x-(sqrt(3)-1))^2+(y-(sqrt(3)-1))^2=(sqrt(3)-1)^2}}}
{{{(x-sqrt(3)+1)^2+(y-sqrt(3)+1)^2=4-2sqrt(3)}}}