Find the equation to the circle which: touches the axes of coordinates and also the line x/a+y/b=1 the centre being in the positive quarant.

Let (h,h) be the center as its center is equidistant from the axes. So rad = h

Now perpendicular distance from center to linex/a+y/b=1 equals radius

h/a+h/b-1 / root(1/a^2 + 1/b^2 ) = h

From here find h

Hence we get the eqn of circle as (x-h)^2 + (y-h)^2 = h^2

thanks