Equation of a Chord Joining Two Points with Circle in Parametric Form
The locus of ...
Question
The locus of midpoint of chord of the circle x2+y2−2x−2y−2=0, which makes an angle of 120∘ at the centre, is
A
x2+y2−x−y+3=0
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B
x2+y2−2x−2y+4=0
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C
x2+y2−4x−4y+8=0
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D
x2+y2−2x−2y+1=0
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Solution
The correct option is Dx2+y2−2x−2y+1=0 For the given circle x2+y2−2x−2y−2=0 Centre, C≡(1,1) and radius, r=√12+12−(−2)=2 From the above figure in △AOM cos60°=OMr⇒OM=1 But OM=√(h−1)2+(k−1)2 ∴(h−1)2+(k−1)2=1 Hence, the locus of (h,k) is (x−1)2+(y−1)2=1 or, x2+y2−2x−2y+1=0.