Circle And System Of Circles Question 242
Question: One of the limit point of the coaxial system of circles containing $ x^{2}+y^{2}-6x-6y+4=0 $ , $ x^{2}+y^{2}-2x $ $ -4y+3=0 $ is
[EAMCET 1987]
Options:
A) $ (-1,,1) $
B) $ (-1,,2) $
C) $ (-2,,1) $
D) $ (-2,,2) $
Show Answer
Answer:
Correct Answer: A
Solution:
Trick: The equation of radical axis is $ S_1-S_2=0 $ i.e., $ 4x+2y-1=0 $ .
$ \therefore $ The equation of circle of co-axial system can be taken as $ (x^{2}+y^{2}-6x-6y+4)+\lambda (4x+2y-1)=0 $ or $ x^{2}+y^{2}-(6-4\lambda )x-(6-2\lambda )y+(4-\lambda )=0 $ ………….(i) whose centre is $ C(3-2\lambda ,\ 3-\lambda ) $ and radius is $ r=\sqrt{{{(3-2\lambda )}^{2}}+{{(3-\lambda )}^{2}}-(4-\lambda )} $ If $ r=0 $ , then we get $ \lambda =2 $ or $ 7/5 $ .
Putting the co-ordinates of C, the limit points are (-1, 1) and $ ( \frac{1}{5},\ \frac{8}{5} ) $ . One of these limit points is given in (a).
BETA
