Conic Sections Question 231
Question: A line is drawn through a fixed point $ P(\alpha ,\beta ) $ to cut the circle $ x^{2}+y^{2}=a^{2} $ at A and B, then PA.PB is equal to
Options:
A) $ {{\alpha }^{2}}+{{\beta }^{2}} $
B) $ {{\alpha }^{2}}+{{\beta }^{2}}-{{\alpha }^{2}} $
C) $ {{\alpha }^{2}} $
D) $ {{\alpha }^{2}}+{{\beta }^{2}}+{{\alpha }^{2}} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Any point on the line at a distance at a r from the point $ P(\alpha ,\beta ) $ is $ (\alpha +rcos\theta ,\beta +rsin\theta ) $ If this point lies on $ x^{2}+y^{2}=a^{2}, $ then $ {{\alpha }^{2}}+r^{2}{{\cos }^{2}}\theta +2ar\cos \theta +{{\beta }^{2}}+r^{2}{{\sin }^{2}}\theta +2\beta r\sin \theta =a^{2} $
$ \Rightarrow r^{2}+2r(\alpha cos\theta +\beta sin\theta )+{{\alpha }^{2}}+{{\beta }^{2}}=a^{2} $
$ \Rightarrow r^{2}+2r(\alpha cos\theta +\beta sin\theta )+{{\alpha }^{2}}+{{\beta }^{2}}-a^{2}=0 $ Nor, if $ PA=r_1 $ and $ PB=r_2, $ then $ r_1 $ and $ r_2 $ must be roots of this equation.
$ \therefore PA.PB=r_1.r_2={{\alpha }^{2}}+{{\beta }^{2}}-a^{2} $
BETA
