Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Let the larger circle have centre O and radius R = 5 cm, and the smaller circle have radius r = 3 cm. Let AB be the chord of the larger circle which touches the smaller circle at point P.
Since AB is a tangent to the smaller circle at P, and OP is the radius through the point of contact, we have OP ⊥ AB. Also, from Example 1 in the context, the chord of the larger circle which touches the smaller circle is bisected at the point of contact. Therefore, P is the midpoint of AB, so AP = BP.
In right triangle OAP, using Pythagoras Theorem: OA² = OP² + AP² 5² = 3² + AP² 25 = 9 + AP² AP² = 16 AP = 4 cm
Therefore, AB = 2 × AP = = 8 cm.
The length of the chord is 8 cm.
Explanation
This question applies the property that a tangent is perpendicular to the radius at the point of contact, and the result from Example 1 that such a chord is bisected at the point of contact. The Pythagoras theorem is then used to find half the chord length, which is doubled to get the final answer.
Solution Steps
Step 1: Identify that OP ⊥ AB since AB is tangent to smaller circle at P
Step 2: Apply property that chord is bisected at point of contact, so AP = BP
Step 3: Apply Pythagoras theorem in ΔOAP: OA² = OP² + AP²
Step 4: Substitute values: 5² = 3² + AP², giving AP = 4 cm
Step 5: Calculate chord length: AB = 2 × AP = 8 cm