A) \[(-1,\,1)\]
B) \[(-1,\,2)\]
C) \[(-2,\,1)\]
D) \[(-2,\,2)\]
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 \[\left( \frac{1}{5},\ \frac{8}{5} \right)\]. One of these limit points is given in (a).
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