A line meets the coordinate axes in $A$ and $B.$ $A$ circle is circumscribed about the triangle $OAB.$The distances from the end points $A, B$ of the side AB to the tangent at $O$ are equal to $m$ and $n$ respectively. Then, the diameter of the circle, is

Let the coordinates of $A$and$B$ be $(a, 0)$ and $(0, b)$ respectively. 

Since the triangle $OAB$is a right angled triangle, the centre of 

the circumcircle is the mid point of $AB$ i.e. $(a/2,b/2)$ and the

 radius is equal to $\frac{1}{2}AB=\frac{1}{2}\sqrt{a^2+b^2}$

The equation of the circumcircle is $x^2+y^2-ax-by=0$

The equation of the tangent at the origin to this circle is 

$ax+by=0.$

$\therefore m=\frac{a^2}{\sqrt{a^2+b^2}}$ and $n=\frac{b^2}{\sqrt{a^2+b^2}}$

$\Rightarrow m+n=\sqrt{a^2+b^2}=$ Diameter of the circle.