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
