Question 1106436
Only here summarizing (i.):


Line perpendicular containing given center of circle,
slope  -4/3,
{{{y=-4x/3+9}}}


Intersection of that with the given tangent line {{{3x-4y=11}}}, {{{y=-11/4+3x/4}}}  is at ( 141/25, 37/25 ).


SQUARE of distance from  (6,1) center and ( 141/25, 37/25 )  is
{{{r^2=(6-141/25)^2+(1-37/25)^2}}}
{{{r^2=9/25}}}


Equation of the circle wanted, {{{(x-6)^2+(y-1)^2=9/25}}}