Circle And System Of Circles Question 173
Question: 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
Options:
A) $ {{(\alpha +\beta )}^{2}}-r^{2} $
B) $ {{(\alpha +\beta )}^{2}}-r^{2} $
C) $ {{(\alpha -\beta )}^{2}}+r^{2} $
D) None of these
Show Answer
Answer:
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 parameter representing the directed distance of any point (x, y) on the line from the point $ P(\alpha ,\ \beta ) $ .
Let this line intersect 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 the circle at A and B, then $ PA=k_1 $ and $ PB=k_2 $ .
$ \therefore $ $ PA.PB=k_1k_2= $ Products of roots $ ={{\alpha }^{2}}+{{\beta }^{2}}-2r^{2} $ .
Trick : As we know from figure, $ PA\ .\ PB=PT^{2} $ .
BETA
