If a tangent PT and a line segment PBA is drawn to a circle with centre O. If OL is perpendicular to AB, prove that PA. PB=PT^2.
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Here is ur answer.
construction. join AB
PA.PB. =(PM-AM)(PM+BM)
(As AM=BM)
=(PM-AM)(PM+AM)
=PM^2-AM^2
(OP^2-OM^2)-AM^2
OP^2-(OM^2+AM^2)
=OP^2-OA^2
but OA = OT (radii )
so PA*AB=PT^2
construction. join AB
PA.PB. =(PM-AM)(PM+BM)
(As AM=BM)
=(PM-AM)(PM+AM)
=PM^2-AM^2
(OP^2-OM^2)-AM^2
OP^2-(OM^2+AM^2)
=OP^2-OA^2
but OA = OT (radii )
so PA*AB=PT^2
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