Question

Two concentric circles are of radii $5\text{ cm}$ and $4\text{ cm}$. Find the length of the chord of the larger circle which touches the smaller circle.

Hint:

A perpendicular line from the center of concentric circles to a chord of the outer circle that is tangent to the inner circle always bisects the chord.
The chord length is always:
\[ L = 2\sqrt{R^2 - r^2} \]
where $R$ is the outer radius and $r$ is the inner radius.
Using this directly:
\[ L = 2\sqrt{5^2 - 4^2} = 2\sqrt{9} = 6\text{ cm} \]

Answer 1

Step 1: Understanding the Question:
The topic is Circles, specifically concentric circles and properties of chords and tangents.
Concentric circles are circles that share the same center.
We have a chord of the outer (larger) circle that is tangent to (touches) the inner (smaller) circle.
We need to determine the length of this chord using geometric properties.

Step 2: Key Formula or Approach:
Let $O$ be the common center. Let $AB$ be the chord of the larger circle touching the smaller circle at point $P$.

• Since $AB$ is tangent to the smaller circle, the radius $OP$ is perpendicular to $AB$: $OP \perp AB$.

• In the larger circle, $OP$ is a perpendicular from the center to the chord $AB$.

• By circle theorems, a perpendicular from the center of a circle to a chord bisects the chord. Hence, $AP = PB$.

This allows us to form a right-angled triangle $OPA$ and use Pythagoras' Theorem:
\[ OA^2 = OP^2 + AP^2 \]

Step 3: Detailed Explanation:

• Let $O$ be the center of the concentric circles.
Let $AB$ be the chord of the larger circle of radius $R = 5\text{ cm}$ which touches the smaller circle of radius $r = 4\text{ cm}$ at point $P$.

• Draw the radius $OP$ to the point of contact $P$ and join $OA$.
$OA$ represents the radius of the larger circle, so $OA = 5\text{ cm}$.
$OP$ represents the radius of the smaller circle, so $OP = 4\text{ cm}$.

• Since $AB$ is a tangent to the inner circle at $P$, the radius $OP$ is perpendicular to the tangent $AB$:
\[ \angle OPA = 90^\circ \]

• In the right-angled triangle $OPA$, apply Pythagoras' Theorem:
\[ OA^2 = OP^2 + AP^2 \]
Substitute the known values:
\[ 5^2 = 4^2 + AP^2 \]
\[ 25 = 16 + AP^2 \]
\[ AP^2 = 25 - 16 = 9 \]
Taking the positive square root:
\[ AP = 3\text{ cm} \]

• Since the perpendicular from the center of a circle to a chord bisects the chord, $P$ is the midpoint of $AB$:
\[ AB = 2 \times AP \]
\[ AB = 2 \times 3 = 6\text{ cm} \]


Step 4: Final Answer:
The length of the chord of the larger circle is $6\text{ cm}$.