1.

A line is drawn through a fix point P(`alpha, beta`) to cut the circle `x^2 + y^2 = r^2` at A and B. Then PA.PB is equal to :

Answer» Point of power= PA*PB=PC*PC=`PT^2`
In`/_PTO`
`OT^2+PT^2=PO^2`
`PT^2=2^2+B^2-r^2`
`PA*PB=PT^2=2^2+B^2=r^2`.


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