Conic-Sections Question 563
Question: The limiting points of the coaxial system determined by the circles $ x^{2}+y^{2}-2x-6y+9=0 $ and $ x^{2}+y^{2}+6x-2y+1=0 $
Options:
A) $ (-1,2),( \frac{3}{5},\frac{-14}{5} ) $
B) $ (-1,2),( \frac{3}{5},\frac{14}{5} ) $
C) $ (-1,2),( \frac{-3}{5},\frac{14}{5} ) $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
[b] The equation of two circles are $ x^{2}+y^{2}-2x-6y+9=0 $ ?. (1) and $ x^{2}+y^{2}+6x-2y+1=0 $ ?. (2) Their radical axis is $ 8x+4y-8=0 $ or $ 2x+y-2=0 $ ? (3) The equation of any circle coaxial with the given circles is $ x^{2}+y^{2}-2x-6y+9+\lambda (2x+y-2)=0 $ or $ x^{2}+y^{2}+(2\lambda -2)x+(\lambda -6)y+(9-2\lambda )=0 $ ?. (4) The centre of this circle is $ [(1-\lambda ),( 3-\frac{\lambda }{2} )] $ Its radius $ =\sqrt{{{(1-\lambda )}^{2}}+{{( 3-\frac{\lambda }{2} )}^{2}}-(9-2\lambda )} $ $ =\sqrt{\frac{5{{\lambda }^{2}}}{4}-3\lambda +1} $ For limiting points its radius = 0 i.e., $ \frac{5{{\lambda }^{2}}}{4}-3\lambda +1=0 $ or $ 5{{\lambda }^{2}}-12\lambda +4=0\therefore $ $ \lambda =2,\frac{2}{5} $ Substituting these values in (5), the limiting points are $ (-1,2) $ and $ ( \frac{3}{5},\frac{14}{5} ) $
BETA
