Question 982783
Let the centre be (x,y),
The distance of centre from all three sides will be equal to radius.
So, for first two sides,
{{{(3x-y-5)/(3^2+(-1^2)) = (x+3y-1)/(1^2+3^2)}}}
i.e. {{{x-2y-2 = 0}}}------ (i)
And for 2nd and 3rd sides,
{{{(x+3y-1)/(1^2+3^2) =  (x-3y+7)/(1^2+3^2)}}}
i.e. {{{6y=6}}}
=> y=1----- (ii)
Putting in (i), we get, x=4
So, centre of circle is (4,1)
Now distance of centre from 2nd side = {{{(4+3*1-1)/(1^2+3^2)}}} = 3/5
So, equation of circle is,
{{{(x-4)^2 + (y-1)^2 = (3/5)^2}}}