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

Question

Aamina · Answer

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 S 1 hence PA.PB =x 1 2 +y 1 2 -r 2 = 2 + 2 -r 2 Hence, second option is correct .

ankit singh · Answer

The equation of any line through the point P(α,β)
cosθx−α=sinθy−β=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+β2−r2=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+β2−r2

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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

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

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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

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

2 Answers

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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