Q17 of 31 Page 65

P is any point in a circle with its centre O. If the length of the radius is 5 cm. and OP = 3 cm., then let us determine the least of the chord passing through the point P.


Given: OP = 3cm, Radius = 5cm


Let OA, OC radius of the circle, OM is perpendicular Bisector Passes through the centre O of Chord CD. OP is a Perpendicular Bisector of Chord AB passes through the Centre O.


In OPM


OMP = 900


Using, Pythagoras Theorem


⇒ OP2 = OM2 + PM2


⇒ OP>OM


We have proved in Question 14, That Nearer chord is greater than the other.


It shows us that


AB<CD


Hence The Least chord passes through point P is AB, OP is Perpendicular bisector of the chord.


In OPA


Using, Pythagoras Theorem


⇒ OA2 = OP2 + AP2


⇒ 52 = 32 + AP2


⇒ AP2 = 25-9


⇒ AP2 = 16


⇒ AP = 4cm



Using above Theorem


AB = 8cm


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