Let be any point within a given circle with centre . Show that the shortest chord of the circle that passes through point is the one that is perpendicular to .

Question

Homi · Accepted Answer

Understand the Question

We are given two concentric circles with a common centre $O$ .
Let $R$ be the radius of the larger circle and $r$ be the radius of the smaller circle.
Chord $BC$ of length $l$ in the larger circle touches the smaller circle at $A$ , making $OA \perp BC$ and bisecting $BC$ such that $AB = \dfrac{l}{2}$ .
The area of the region between the two concentric circles is given by $\pi R^2 - \pi r^2 = \pi (R^2 - r^2)$ .
We use Pythagoras theorem in $\Delta OAB$ to express $R^2 - r^2$ in terms of $l$ .

Step 1 · Relate Radii and Chord using Pythagoras Theorem

Let the radius of the larger circle be $R$ ( $OB = R$ ) and the radius of the smaller circle be $r$ ( $OA = r$ ). Diagram 1

Since the radius is perpendicular to the tangent at the point of contact, $OA \perp BC$ . The perpendicular from the centre to a chord bisects the chord: $AB = \dfrac{BC}{2} = \dfrac{l}{2}$

In right-angled triangle $OAB$ , by Pythagoras theorem: $OA^2 + AB^2 = OB^2$

\begin{aligned} r^2 + \left(\dfrac{l}{2}\right)^2 &= R^2 \\[0.6em] r^2 + \dfrac{l^2}{4} &= R^2 \end{aligned}

$R^2 - r^2 = \dfrac{l^2}{4}$

Step 2 · Calculate the Area of the Enclosed Region

The area enclosed between the two concentric circles is: $\text{Area} = \pi R^2 - \pi r^2 = \pi (R^2 - r^2)$

Substituting $R^2 - r^2 = \dfrac{l^2}{4}$ : $\text{Area} = \pi \left(\dfrac{l^2}{4}\right) = \dfrac{1}{4} \pi l^2$

Answer

$\dfrac{1}{4} \pi l^2$

Common Mistakes

Trying to find $R$ and $r$ individually: You do not need the individual values of $R$ and $r$ ; only the difference of their squares ( $R^2 - r^2$ ) is required to find the area.
Incorrect chord segment length: Forgetting to divide the chord length by $2$ , mistakenly using $AB = l$ instead of $AB = \dfrac{l}{2}$ in the Pythagoras equation.

Question 23

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