question_answer

The locus of midpoint of the chords of the circle \[{{x}^{2}}+{{y}^{2}}-2x-2y-2=0\]which makes an angle of \[120{}^\circ \] at the centre is [MNR 1994]

A)            \[{{x}^{2}}+{{y}^{2}}-2x-2y+1=0\]                              

B)            \[{{x}^{2}}+{{y}^{2}}+x+y-1=0\]

C)            \[{{x}^{2}}+{{y}^{2}}-2x-2y-1=0\]                                

D)            None of these

Correct Answer: A

Solution :

           The centre of given circle is (1, 1) and its radius is \[\sqrt{2}\]. From the figure, if \[M(h,\ k)\] be the middle point of chord AB subtending an angle \[\frac{2\pi }{3}\] at C, then                      \[\frac{CM}{AC}=\cos \frac{\pi }{3}=\frac{1}{2}\Rightarrow 4C{{M}^{2}}=A{{C}^{2}}\]                    or \[4[{{(h-1)}^{2}}+{{(k-1)}^{2}}]=4\Rightarrow {{h}^{2}}+{{k}^{2}}-2h-2k+2=1\]                    Hence the locus is\[{{x}^{2}}+{{y}^{2}}-2x-2y+1=0\].