A) \[x+y-7=0\]
B) \[2x-y-5=0\]
C) \[~x+2y-3=0\]
D) none of these
Correct Answer: C
Solution :
Key Idea: The perpendicular distance from centre to the chord its bisects the chord. Given equation of circle is \[{{x}^{2}}+{{y}^{2}}+2x-4y-11=0\] \[\therefore \] Centre is, \[(-1,2).\] Since, the diameter of the circle is perpendicular to the chord of the circle. Hence, the equation of diameter is \[x+2y+\lambda =0\] ...(i) Its passes through the centre \[(-1,2)\] of the circle \[\Rightarrow \] \[-1+4+\lambda =0\]\[\Rightarrow \]\[\lambda =-3\] Putting the value of \[\lambda \] in Eq. (i), we get \[x+2y-3=0\]
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