Question

The equations of the circles which touched both the axes and also the straight line 4x + 3y = 6 in the first quadrant and lies below it is $4x^2+4y^2-$ -4x - 4y + 1 = 0

Answer 1

Circle touching both axes is (h, h), h i.e.
$(x-h{)}^2+(y-h{)}^2=h^2$
The condition of tangency with given line gives$\frac{4h+3h-6}{\pm 5}=hor7h-6=5h$
h = 3, 1/2
Since the circle is to lie below the line, the value of h cannot exceed 3/2 hence h = 3 is rejected.
$\therefore$ h = 1 / 2.
Putting in (1) we get the answer.