TITLE: Maximizing an angle based on certain constraints
QUESTION [10 upvotes]: $A (0,a)$ and $B(0,b)\; (a,b>0)\;$ are the vertices of $\triangle ABC$ where $C(x,0)$ is variable. Find the value of $x$ when angle $ACB$ is maximum.

Now geometry's never really been my strong point, so I decided to go with a bit of calculus.
First, I used the sine rule:
$$\mathrm{sinC=\frac{b-a}{2R}} $$
where R is the radius of the circumcircle. I note that for angle C to be maximum, sinC should be maximum. As such, R must be minimum.
Next, I used the relation $$\mathrm{R=\frac{(b-a)\cdot\sqrt{x^2+b^2}\cdot\sqrt{x^2+a^2}}{2\Delta}} $$
where $\mathrm{\Delta \text{ is the area of }ABC=\frac{(b-a)x}{2}}$.
A bit of comparatively lengthy differentiation gives me the value of $x$ as $\sqrt{ab}$.
When I go through the solutions, it's simply been stated:

For angle ACB to be maximum, the circle passing through A,B will touch the X-axis at C.

Beyond this, it's been solved using the very simple $\mathrm{OC^2=OA\cdot OB}$, where O is the origin. So the above statement seems to be the difference between a lengthy differentiation and a one line solution.
It's getting a little difficult for me to see why the above statement should be intuitive. Could someone shed a bit more light on it for me, and possibly provide an intuitive proof?

REPLY [2 votes]: There are two ways of looking at it.
Circles such as those that pass through non red sides meeting on x-axis at D,E subtend the same angle from segment AB ( Angles in a segment are equal property). In order that there be a unique point, these points should be drawn together to make them into a repeated point. A repeated point is in fact a point of tangentcy at C.
By means of the Circle's property that the product of segments be constant (this being is the power of the Circle ) we have
$$ OA \cdot OB= OC^2= x^2 \to \; x = \sqrt {ab} \tag1$$

Next way is direct confirmation with differential calculus, maxima/minima.
The "look angle " or subtended angle is
$$ \tan^{-1}\frac{a}{x}-\tan^{-1}\frac{b}{x} $$
Differentiate w.r.t. $x$ arctan and Chain Rule
$$ \dfrac{-a/x^2}{1+a^2/x^2} + \dfrac{-b/x^2}{1+b^2/x^2} =0$$
When simplified, we get the same result as (1).