InterviewSolution
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InterviewSolution
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A line is drawn through a fixed point `P(alpha, B)` to cut the circle `x^(2)+y^(2)=r^(2)` at A and B. Then PA.PB is equal toA. `(alpha+beta)^(2)-r^(2)`B. `(alpha^(2)+beta^(2)-r^(2)`C. `(alpha-beta)^(2)+r^(2)`D. None of the above |
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Answer» Correct Answer - B The equation of any line through the point `P(alpha,beta)` is `(x-alpha)/(cos theta)=(y-beta)/(sin theta)=k("say")` Any point on this line is `(alpha+k cos theta, beta+k sin theta)` This point lies on the given circle, if `(alpha+k cos theta)^(2)+(beta+k sin theta)^(2)=r^(2) or k^(2)+2k(alpha cos theta+beta sin theta)+alpha^(2)+beta^(2)-r^(2)=0......(i)` Which being quadratic in k, gives two values of k. Let PA=`k_(1),PB=k_(2)`, where `k_(1),k_(2)` are the roots of Eq. (i) then, `PA.PB=k_(1)k_(2)=alpha^(2)+beta^(2)-r^(2)` |
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