In the following figure, if PT = 6 and AB = 5, find PB?
Explanation:
Let PA = x
When a secant and a tangent are drawn from an external point to a circle
⇒ PA × PB = PT2
∴ x × (x + 5) = 62
⇒ x2 + 5x – 36 = 0
⇒ (x + 9)(x - 4) = 0
∴ x = -9 or 4. (x = -9 is rejected)
∴ PB = x + 5 = 9.
Hence, 9.
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