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, \[PA\ .\ PB=P{{T}^{2}}\].You need to login to perform this action.
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