in the above given below figure from an external point P ,a tangent PT and a secant PAB is drawn to circle with center O.ON is perpendicular on chord AB.Prove that PA.PB=PT2
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AFTER THAT U HAVE TO WRITE THIS
PA. PB=(PM-AM) (PA+AM)
=PM^2-AM^2
=(OP^2-OM^2)-(OA^2-OM^2)
=OP^2-OA^2
=OP^2-OT^2 [SINCE OA=OT]
= PT^2 [SINCE PT IS PERPENDICULAR TO OT ] RADIUS OF THE CICLE
THEREFORE PA. PB=PT^2
PA. PB=(PM-AM) (PA+AM)
=PM^2-AM^2
=(OP^2-OM^2)-(OA^2-OM^2)
=OP^2-OA^2
=OP^2-OT^2 [SINCE OA=OT]
= PT^2 [SINCE PT IS PERPENDICULAR TO OT ] RADIUS OF THE CICLE
THEREFORE PA. PB=PT^2
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