P(a, b) is a point in the first quadrant. Circles are drawn through P touching the coordinate axes, such that the length of common chord of these circles is maximum, find the ratio a : b?

In order for a circle to be touching both coordinate axes in the first quadrant, it must have its center located on the line #y=x#, as seen below:

The common chord is going to be #PQ#, where #Q# is the symmetric point (a.k.a. the reflection) of #P# with respect to the line #y=x#, hence #Q# has coordinates #(b,a)#.

In order for the common chord to be the maximised, it has to be the diameter of the circle where #P# and #Q# are on its "high ends":

In our particular example, it almost overlaid one of the original circles.

Anyway, the center of this new circle is the midpoint of #PQ#, which we will denote #L#.

#L((a+b)/2,(a+b)/2)#

The equation of the circle is:

#(x-(a+b)/2)^2+(y-(a+b)/2)^2 = ((PQ)/2)^2#

The lenght of #PQ# is, as calculated by you, #sqrt2|a-b|#.

After some arithmetic, we reach the equation

#(x−a)(x−b)+(y−a)(y−b)=0#

The circle with diameter #PQ# must touch the axes as well, hence the points #((a+b)"/"2,0)# and #(0,(a+b)"/"2)# must be on it as well.

If we substitute the coordinates of one of these points (doesn't matter which; it gives us the same answer) into the circle equation we get the condition

#(a+b)^2=8ab#

#a^2+2ab+b^2=8ab#

#a^2-6ab+b^2=0#

Divide both sides by #b^2#:

#a^2/b^2 - 6a/b + 1 =0#

#(a/b)^2 - 6(a/b) + 1 =0#

This is a quadratic equation with the indeterminate the ratio of #a# and #b#, exactly what we want!

The solutions of this equation are

#a:b = 3+-2sqrt2#

These are the ratios we desire.