Question 1066857
The problem, as stated, has an infinite number of solutions.  

I will find one circle then explain why there are an infinite number of solutions.  
The circle I will find is the one tangent to the line 3y-4x-11=0  AND also tangent to a line parallel to this line where this 2nd line passes through (8,4).  The circle is tangent to this 2nd line at (8,4).   In this way, there is only one solution. 

{{{3y - 4x - 11 = 0 }}}  —>   {{{ y = (4/3)x + (11/3) }}}   

Now imagine a line passing through (8,4), then the center of the circle, then through {{{ y=(4/3)x+(11/3) }}} at a point where the line is tangent to the circle.  This (3rd) line has slope -3/4 because it must be perpendicular to  {{{ y = (4/3)x + (11/3) }}}   (if a line has slope m then a line perpendicular to it has slope -1/m).

The line through the center of the circle also passes through (8,4)  so we can find the equation of it.

                       y = mx+b
                       4 = (-3/4)(8) + b
                       4 + (3/4)(8) = b
                       4 + 6 = b  —> b=10
        
                  {{{ y = (-3/4)x + 10 }}}    <<< line through center of circle, 
                                                                 perpendicular to  {{{ y = (4/3)x + (11/3) }}}   

Where these two lines meet, their y values are the same:

                        {{{ (-3/4)x + 10 }}} = {{{ (4/3)x + (11/3) }}}

                         This reduces to  {{{ x = 76/25 }}}  —>  {{{ y = 193/25 }}}

So the diameter of the circle is:

 {{{ sqrt((8 - 76/25)^2 + (4 - 193/25)^2)) }}}  = {{{ 6.20 }}}  ——>   {{{ r = 3.10 }}}

The center of the circle is at  ( {{{ (8+(75/25))/2 }}} ,  {{{  (4+(193/25))/2  }}} )
   which works out to  ( {{{ 5.52 }}}, {{{ 5.86}}} )
—

The equation of the circle is therefore:   {{{ highlight( (x-5.52)^2 + (y-5.86)^2 = (3.10)^2 )}}}

—

Now, the original problem statement has an infinite number of solutions because you can make a circle tangent to   {{{ y = (4/3)x + (11/3) }}}   that is a little bigger than the one we just found above, and passes through (8,4) but at a different point on the circle than the line perpendicular to  {{{ y = (4/3)x + (11/3) }}}.   This can be done an infinite number of times by moving the tangent point (& adjusting circle size) by an infinitesimally small amount.
—

{{{ drawing(300,300, -10,10,-10,10,
                grid(1),
                green(circle(5.52, 5.56,3.10) ),
                line(-10,-9.6667,  10, 17),
                line(-10, 17.5, 10, 2.5),
                line(0, -20/3, 9, 5.333)
     )
}}}