A) \[{{x}^{2}}+{{y}^{2}}+3x+3y+8=0\]
B) \[{{x}^{2}}+{{y}^{2}}-3x-3y-8=0\]
C) \[{{x}^{2}}+{{y}^{2}}-3x+3y+8=0\]
D) None of these
Correct Answer: B
Solution :
Let us find the equation of family of circles through \[(2,\ -2)\] and \[(-1,\ -1)\]. i.e. \[(x-2)(x+1)+(y+2)(y+1)+\lambda \left( \frac{y+2}{-2+1}-\frac{x-2}{2+1} \right)=0\] Now for point (5, 2) to lie on it, we should have \[\lambda \]given by \[3\ .\ 6+4\ .\ 3+\lambda \left( \frac{4}{-1}-1 \right)=0\Rightarrow \lambda =\frac{30}{5}=6\] Hence equation is \[(x-2)(x+1)+(y+2)(y+1)+6\left( \frac{y+2}{-1}-\frac{x-2}{3} \right)=0\] or \[{{x}^{2}}+{{y}^{2}}-3x-3y-8=0\]. Trick: Here the circle \[{{x}^{2}}+{{y}^{2}}-3x-3y-8=0\]is satisfied by (2, ?2), (?1, ?1) and (5, 2). Therefore students must check such type of problems conversely.You need to login to perform this action.
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