The locus of the mid-points of the chords of | vedclass

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(C) Given,the equation of the circle is $x^{2}+y^{2}+2x-2y-2=0$.This can be rewritten as $(x+1)^{2}+(y-1)^{2}=4$.Thus,the centre is $(-1, 1)$ and the

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$x^{2}+y^{2}+2x-2y=0$

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(C) Given,the equation of the circle is $x^{2}+y^{2}+2x-2y-2=0$.This can be rewritten as $(x+1)^{2}+(y-1)^{2}=4$.Thus,the centre is $(-1, 1)$ and the radius $r = 2$.Let $P(h, k)$ be the mid-point of a chord $AB$.The distance $OP$ from the centre $O(-1, 1)$ to the mid-point $P(h, k)$ is given by $OP = \sqrt{(h+1)^{2}+(k-1)^{2}}$.Since the chord $AB$ subtends an angle of $90^{\circ}$ at the centre,$	riangle OAP$ is a right-angled triangle with $\angle OAP = 45^{\circ}$.In $	riangle OAP$,$\sin 45^{\circ} = \frac{OP}{OA}$.$\frac{1}{\sqrt{2}} = \frac{\sqrt{(h+1)^{2}+(k-1)^{2}}}{2}$.Squaring both sides,we get $\frac{1}{2} = \frac{(h+1)^{2}+(k-1)^{2}}{4}$.$(h+1)^{2}+(k-1)^{2} = 2$.Expanding this,$h^{2}+2h+1+k^{2}-2k+1 = 2$.$h^{2}+k^{2}+2h-2k = 0$.Replacing $(h, k)$ with $(x, y)$,the locus is $x^{2}+y^{2}+2x-2y=0$.

The locus of the mid-points of the chords of the circle $x^{2}+y^{2}+2x-2y-2=0$ which make an angle of $90^{\circ}$ at the centre is

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