Question 936629
First find the equation of the circle by completing the square in x and y.
{{{x^2-2x+y^2-2y=-1}}}
{{{(x^2-2x+1)+(y^2-2y+1)=-1+1+1}}}
{{{(x-1)^2+(y-1)^2=1}}}
The circle is centered at (1,1) and has a radius of 1.

{{{drawing(300,300,-2,5,-2,5,green(line(1.464,1.464,5,5)),blue(line(0,.89,.89,.89)),blue(line(0.89,.89,1.464,1.464)),line(0,0,0,10),line(0,0,10,0),locate(0.3,0.9,r),locate(1.2,1.3,r),circle(1.49,1.49,0.1),circle(0.89,.89,.89),locate(3,3,R=1),(circle(5,5,5)))}}}
The point of intersection of the large circle and the small circle is shown in the diagram. 
The coordinates of the position are given by,
{{{x=y=1-1(sqrt(2)/2))=1-sqrt(2)/2))}}}
or approximately (0.2929,0.2929).
That distance is equal to,
{{{r+(sqrt(2)/2)r=1-sqrt(2)/2}}}
{{{r=(1-sqrt(2)/2)/(1+sqrt(2)/2)}}}
{{{r=0.1715}}}
So the smaller circle is centered at (0.1715,0.1715) and has a radius of 0.1715.
So the equation would be,
{{{(x-0.1715)^2+(y-0.1715)^2=0.1715^2}}}
{{{highlight((x-0.1715)^2+(y-0.1715)^2=0.0294)}}}