I know what happens because i played around with software, but i cant see an easy way why?
coordinates:
$\displaystyle A\left(a,\sqrt{a(2-a)}\right)$
and
$\displaystyle B\left(0,\sqrt{2a}\right)$
where $\displaystyle 0<a<1$
write down the equation of line AB and find its intersection with the x-axis
$\displaystyle C\left(2+\sqrt{4-2a},0\right)$
take the limit as $\displaystyle a\to 0$
Now, I cant see how you got a neat form of the x coord of C! I get a very messy expression..
Ok ..so you rationalised a surd. Is there a Geometric way of looking at this? If I change the circle the x coord of C appears to be double the diameter.
If AH is perpendicular to OC, H on OC, we can use similar triangles to get
$\displaystyle \frac{\text{CH}}{\text{CO}}=\frac{\text{AH}}{\text {BO}}$
$\displaystyle \frac{x-a}{x}=\frac{\sqrt{a(2-a)}}{\sqrt{2a}}$
Solve for x
$\displaystyle x=\frac{2 a}{2 - \sqrt{4-2a }}$
rationalize