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प्रश्न
In fig., PT is a tangent to the circle at T and PAB is a secant to the same circle. If PB = 9cm and AB = 5 cm, find PT.

योग
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उत्तर
Given AP and AQ are diameters of circles with centre O and O1 respectively
∴ ∠ APB = 90° ---( 1) (Angle in a semidrde is a right angle)
Similarly, ∠ ABQ = 90° ---(2)
Adding (1) and (2)
∠ APB + ∠ ABQ = 90° + 90°
∠ PBQ = 180°
Hence, PBQ is a straight line
∴ P, B and Q are collinear.
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