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.
You will be redirected in
3 sec
