Circle And System Of Circles Question 293
Question: 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]
Options:
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
Show Answer
Answer:
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 4CM^{2}=AC^{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 $ .
BETA
