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

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(C) The equation of the circle is $x^2 + y^2 - 2x - 4y - 11 = 0$.Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get the centre $C = (1, 2)$ and rad

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$x^2 + y^2 - 2x - 4y - 7 = 0$

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(C) The equation of the circle is $x^2 + y^2 - 2x - 4y - 11 = 0$.Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get the centre $C = (1, 2)$ and radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{1^2 + 2^2 - (-11)} = \sqrt{1 + 4 + 11} = \sqrt{16} = 4$.Let $(h, k)$ be the mid-point of a chord. The distance $d$ from the centre $(1, 2)$ to the chord is given by $d = \sqrt{(h-1)^2 + (k-2)^2}$.In the right-angled triangle formed by the centre,the mid-point,and an endpoint of the chord,the angle at the centre is $60^o / 2 = 30^o$.Thus,$\cos 30^o = \frac{d}{r} = \frac{\sqrt{(h-1)^2 + (k-2)^2}}{4}$.Since $\cos 30^o = \frac{\sqrt{3}}{2}$,we have $\frac{\sqrt{(h-1)^2 + (k-2)^2}}{4} = \frac{\sqrt{3}}{2}$.Squaring both sides,$\frac{(h-1)^2 + (k-2)^2}{16} = \frac{3}{4}$.$(h-1)^2 + (k-2)^2 = 12$.$h^2 - 2h + 1 + k^2 - 4k + 4 = 12$.$h^2 + k^2 - 2h - 4k - 7 = 0$.Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 - 2x - 4y - 7 = 0$.

The locus of the mid-points of the chords of the circle $x^2 + y^2 - 2x - 4y - 11 = 0$ which subtend $60^o$ at the centre is

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