WBJEE · Maths · Circle
The locus of the mid-points of the chords of the circle \(x^{2}+y^{2}+2 x-2 y-2=0,\) which make an angle of \(90^{\circ}\) at the centre is
- A \(x^{2}+y^{2}-2 x-2 y=0\)
- B \(x^{2}+y^{2}-2 x+2 y=0\)
- C \(x^{2}+y^{2}+2 x-2 y=0\)
- D \(x^{2}+y^{2}+2 x-2 y-1=0\)
Answer & Solution
Correct Answer
(C) \(x^{2}+y^{2}+2 x-2 y=0\)
Step-by-step Solution
Detailed explanation
Given, equation of circle is \(x^{2}+y^{2}+2 x-2 y-2=0\) \(\Rightarrow \quad(x+1)^{2}+(y-1)^{2}=4\) \(\therefore\) Centre (-1,1) and radius \(=2\) Let \((h, k)\) be the mid-point of chord. From figure, \[ O P=\sqrt{(h+1)^{2}+(k-1)^{2}} \] \(\ln \Delta O A P\)…
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