question_answer

A line is drawn through a fixed point \[P(\alpha ,\ \beta )\] to cut the circle \[{{x}^{2}}+{{y}^{2}}={{r}^{2}}\] at A and B. Then \[PA\ .\ PB\] is equal to

A)            \[{{(\alpha +\beta )}^{2}}-{{r}^{2}}\]                               

B)            \[{{(\alpha +\beta )}^{2}}-{{r}^{2}}\]

C)            \[{{(\alpha -\beta )}^{2}}+{{r}^{2}}\]                               

D)            None of these

Correct Answer: B

Solution :

           Let the equation of line through the point \[(\alpha ,\ \beta )\] be \[\frac{x-\alpha }{\cos \theta }=\frac{y-\beta }{\sin \theta }=k\] (say)                                   ?.(i)                    where k is the distance of any point (x, y) on the line from the point \[P(\alpha ,\ \beta )\]. Let this line meets the circle \[{{x}^{2}}+{{y}^{2}}={{r}^{2}}\] at \[(\alpha +k\cos \theta ,\ \beta +k\sin \theta )\].                    \[\therefore \]\[{{(\alpha +k\cos \theta )}^{2}}+{{(\beta +k\sin \theta )}^{2}}={{r}^{2}}\]                    or \[{{k}^{2}}+2(\alpha \cos \theta +\beta \sin \theta )k+({{\alpha }^{2}}+{{\beta }^{2}}-{{r}^{2}})=0\],                    which is a quadratic in k. If \[{{k}_{1}}\] and \[{{k}_{2}}\] are its roots and the line (i) meets circle at A and B, then \[PA={{k}_{1}}\] and \[PB={{k}_{2}}\].                    \[\therefore \]\[PA.PB={{k}_{1}}{{k}_{2}}=\]Products of roots\[={{\alpha }^{2}}+{{\beta }^{2}}-{{r}^{2}}\].                    Trick : As we know from figure, \[PA\ .\ PB=P{{T}^{2}}\].