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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


4 years ago

							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 S1hence PA.PB =x12+y12-r2                      = 2+2-r2          Hence, second option is correct .

4 years ago
							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=r2or k2+2kα+(αcosθ+βsinθ)+α2+β2−r2=0.....(i)Which being quadratic in k, gives two values of kLet PA=k1​, PB=k2​, where k1​,k2​ are teh roots of Eq. (i) thenPA.PB=k1​k2​=α2+β2−r2

one month ago
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