Question

Prove the following : If a secant is drawn from a point P which cuts the circle at A and B, then the product PA.PB= constant.
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Answer 1

Answer:

PA.PB=(PC)

2

where C is the point of contact of a tangent drawn to the circle from point P.

If the center of the circle is O,ΔPOC forms a right-angled triangle where

(PO)

2

=(OC)

2

+(PC)

2

⇒(4−0)

2

+(7−0)

2

=9+(PC)

2

⇒(PC)

2

=16+49−9=56

∴PA.PB=56

solution