Locus of the circumcenter of triangle formed by the axes and tangent to a given circle. 
A circle centered at $(2,2)$ touches the coordinate axes and a straight variable line $AB$ in the first quadrant, such that $A$ lies the $Y-$ axis, $B$ lies of the $X-$ axis and the circle lies between the origin and the line $AB$. Find the locus of the circumcenter of the triangle $OAB$, where $O$ denotes the origin.
Answer: $xy=x+y+\sqrt{x^2+y^2}$

I was able to solve this question with a very lengthy approach outlined below:
Since $\Delta OAB$ is always right angled at $O$, it's circumcenter will be the midpoint of the segment $AB$.
Also, the equation of the given circle would be

$$(x-2)^2+(y-2)^2=4$$

Then I considered the question of the line $AB$ to be

$y+mx=c$, where $m$ is positive number.

Then I used the fact that this line is tangent to the given circle, i.e. the perpendicular distance of the line from the point $(2,2)$ is $2$ units to obtain a quadratic equation in $c$:
$$c^2-4(1+m)c+8m=0$$
Now, this yielded two values for $c$ of which one has to be rejected because in that case the line was trapped between the circle and the origin.
Thus, $c=2+2m+2\sqrt{1+m^2}$.
From here, the coordinates of the circumcenter required are:
$\left(\frac{1+m+\sqrt{1+m^2}}{m},1+m+\sqrt{1+m^2}\right)$.
Luckily, this was on a MCQ test and all the options had the terms $xy$, $x+y$ and $\sqrt{x^2+y^2}$ present. So I could evaluate these values and  then see which option is correct.
But the above method is far too lengthy, and I'm looking for a shorter method, if it exists.
Please note that the average time for a question in the test was nearly two to three minutes.
Thanks a bunch!

The first thing to note is that $\triangle OAB$ is always a right triangle, and its incenter is always $(2,2)$.  Thus, hypotenuse $AB$ is always a diameter of the circumcircle, and the circumcenter is therefore the midpoint of $AB$.  If the line passing through $AB$ has equation $$\frac{x}{a} + \frac{y}{b} = 1,$$ then the $x$-intercept is $(a,0)$, and the $y$-intercept is $(0,b)$.  The area enclosed by the triangle may be calculated in two ways:  $$|\triangle OAB| = \frac{ab}{2} = rs$$ where $r = 2$ is the inradius and $$s = \frac{1}{2}\left(a + b + \sqrt{a^2 + b^2}\right)$$ is the semiperimeter.  Consequently, $$\frac{ab}{2} = a+b+\sqrt{a^2+b^2}.$$  Since $(x,y) = (a/2, b/2)$ is the circumcenter, we obtain the relationship $$2xy = 2x + 2y + \sqrt{(2x)^2+(2y)^2},$$ or $$xy = x + y + \sqrt{x^2 + y^2},$$ as claimed.

The implicit formula for the locus admits a natural parametrization using the angle $\theta$ formed by the ray from the origin to $(x,y)$ and the positive $x$-axis:  $$(x,y) = \left(1 + \tan \left(\frac{\theta}{2} + \frac{\pi}{4}\right), 1 + \cot \frac{\theta}{2} \right), \quad 0 < \theta < \frac{\pi}{2}.$$

