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Chord AB of a circle is product to a poi...

Chord AB of a circle is product to a point P and C is a poimt on the circle such that PC is a tangent to the circle. If PC = 18 cm, and BP = 15 cm, then AB is equal to:

A

8.5 cm

B

6.2 cm

C

5.8 cm

D

6.6 cm

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem step by step, we will use the properties of tangents and chords in a circle. ### Step 1: Understand the given information We have a circle with a chord AB that is extended to a point P outside the circle. There is a point C on the circle such that PC is a tangent to the circle. We are given: - PC = 18 cm (the length of the tangent from point P to point C) - BP = 15 cm (the length from point B to point P) ### Step 2: Apply the tangent-secant theorem According to the tangent-secant theorem, the square of the length of the tangent segment (PC) is equal to the product of the lengths of the entire secant segment (AP) and its external segment (PB). Mathematically, this can be expressed as: \[ PC^2 = PB \times PA \] ### Step 3: Substitute the known values We know: - \( PC = 18 \) cm - \( BP = 15 \) cm Substituting these values into the equation: \[ 18^2 = 15 \times PA \] \[ 324 = 15 \times PA \] ### Step 4: Solve for PA Now, we can solve for \( PA \): \[ PA = \frac{324}{15} \] \[ PA = 21.6 \text{ cm} \] ### Step 5: Relate PA to AB and PB From the relationship of the segments, we know: \[ PA = PB + AB \] Substituting the known values: \[ 21.6 = 15 + AB \] ### Step 6: Solve for AB Now we can find the length of AB: \[ AB = 21.6 - 15 \] \[ AB = 6.6 \text{ cm} \] ### Final Answer Thus, the length of chord AB is **6.6 cm**. ---
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