Depends On The Circle, Right?

It is indeed true that the length of the largest chord possible in a circle is always equal to twice its radius.

PROOF:

Let a circle with center $O$ have a chord $AB$ . Join $OA$ and $OB$ . $P$ is a point on $AB$ such that $OP⊥AB$ .

We know that the perpendicular from the centre of a circle to a chord bisects the chord. Then $PA=PB= \dfrac12 AB$ .

Let the radius = $r$ , chord AB length = $c$ and perpendicular distance between centre and the chord $AB = x$ .

Now by Pythagorean theorem , in $\triangle OPA$ ,
$OA^2 =OP^2 +AP^2$ .
$r^2 = x^2 + \left( \dfrac c2\right)^2$ .

By making $c$ the subject of he formula, $c = 2\sqrt{r^2-x^2}$ .

So, we need to find the maximum possible value of $c$ , right? We know that a function is maximized when its derivative is equal to zero. Then, since radius $r$ is constant, $c$ is maximum when:

$\begin{aligned} \dfrac{dc}{dx} &=& 0 \\ \Rightarrow \dfrac d{dx} \left (2\sqrt{r^2-x^2} \right) &=& 0 \\ \Rightarrow 2 \dfrac d{dx} \left (\sqrt{r^2-x^2} \right ) &=& 0 \\ \Rightarrow 2 \left( \dfrac x{\sqrt{r^2-x^2}} \right) &=& 0 \\ \Rightarrow x &=& 0 \end{aligned}$

Then, $c$ is maximum when $x=0$ . Then clearly the maximum value of $c$ is:

$c = 2\sqrt{r^2 - 0^2} = 2\sqrt{r^2} = 2r = 2 \times \text{ radius }.$

The diameter is known as the largest chord and we know that $d = 2r$ .