1. ## Circle problem

I know what happens because i played around with software, but i cant see an easy way why?

2. ## Re: Circle problem

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 $\displaystyle a\to 0$

3. ## Re: Circle problem

why is the y coordinate of B sqrt(2a) ?

Ok ignore ..use the property that OA=OB ..

Thanks

4. ## Re: Circle problem

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.

5. ## Re: Circle problem

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