A secant PAB is drawn from an external point P to the circle with the centre at O, intersecting it at A and B. If OP = 17 cm, PA = 12 cm and PB = 22.5 cm, then what is the radius of the circle?

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Given: OP = 17 cm, PA = 12 cm, PB = 22.5 cm.

Since PAB is a secant, the power of point P with respect to the circle states:
PA × PB = PT², where T is the tangent point if a tangent is drawn from P.

Calculate PA × PB:
PA × PB = 12 × 22.5 = 270 cm².

Let the radius of the circle be r.

Using the power of a point theorem:
PT² = PA × PB = 270.

Also, OP = 17 cm, and PT is the length of the tangent from P to the circle.

By Pythagoras theorem in triangle OPT (O is center, T is tangent point):
OP² = OT² + PT²
=> 17² = r² + 270
=> 289 = r² + 270
=> r² = 289 - 270 = 19

Therefore, radius r = √19 cm.

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