The eqn of circle which touches both the axes and straight line 4x+3y=6 in the first quadrant and lies below it is.

The given line cuts the co-ordinate Axes at (3/2, 0) and (0, 2). This implies the circle lies in the first quadrant. The circle is equidistant from both the Axes thus its equation is of the form: (x-a)²+(y-a)²=a², where the distance of the centre of the circle from the axes and the given line ie. 4x+3y=6,  and the radius of the cirle is 'a' units. Coordinate of centre (a,a) and it is at a distance of a unit from given line. 

So, (4a+3a-6)/√(4²+3²) = a

   or, (7a-6)/5=a

   or, 7a-6=5a

   or, 2a=6

    So, a=3

 Therefore equation of the required circle is (x-3)²+(y-3)²=3²