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Grade: 12

                        

A line is drawn through a fixed point P(α,β) to cut the circle x^2+y^2=r^2 at A and B. Then PA.PB is (α+β)^2-r^2 α^2+β^2-r^2 (α-β)^2+r^2 None of these

4 years ago

Answers : (2)

Aamina
64 Points
							
when a line through a fixed point P cuts the circle in two different points A and B then PA.PB is called power of the point which is equal to S1
hence 
PA.PB =x12+y12-r2
                      = \small \alpha2+\small \beta2-r2      
    Hence, second option is correct .
4 years ago
ankit singh
522 Points
							

The equation of any line through the point P(α,β)
cosθxα=sinθyβ=k(say)
Any point on this line is
(α+kcosθ,β+ksinθ)
This point lies on the given circle, if
(α+kcosθ)2+(β+ksinθ)2=r2
or k2+2kα+(αcosθ+βsinθ)+α2+β2r2=0.....(i)
Which being quadratic in k, gives two values of k
Let PA=k1, PB=k2, where k1,k2 are teh roots of Eq. (i) then
PA.PB=k1k2=α2+β2r2

one month ago
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