The equation of the locus of the midpoints of | vedclass

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(A) The given circle is $4x^2 + 4y^2 - 12x + 4y + 1 = 0$,which can be written as $x^2 + y^2 - 3x + y + \frac{1}{4} = 0$.Comparing with $x^2 + y^2 + 2g

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$16(x^2 + y^2) - 48x + 16y + 31 = 0$

Vedclass · Answer

(A) The given circle is $4x^2 + 4y^2 - 12x + 4y + 1 = 0$,which can be written as $x^2 + y^2 - 3x + y + \frac{1}{4} = 0$.Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get center $C = (\frac{3}{2}, -\frac{1}{2})$ and radius $r = \sqrt{(\frac{3}{2})^2 + (-\frac{1}{2})^2 - \frac{1}{4}} = \sqrt{\frac{9}{4} + \frac{1}{4} - \frac{1}{4}} = \frac{3}{2}$.Let $M(h, k)$ be the midpoint of a chord. The distance $d$ from the center $C$ to the chord is $d = \sqrt{(h - \frac{3}{2})^2 + (k + \frac{1}{2})^2}$.The chord subtends an angle of $\frac{2\pi}{3}$ at the center,so in the right-angled triangle formed by the center,the midpoint,and an endpoint of the chord,the angle at the center is $\frac{\pi}{3}$.Thus,$d = r \cos(\frac{\pi}{3}) = \frac{3}{2} 	imes \frac{1}{2} = \frac{3}{4}$.Squaring both sides: $(h - \frac{3}{2})^2 + (k + \frac{1}{2})^2 = (\frac{3}{4})^2$.$h^2 - 3h + \frac{9}{4} + k^2 + k + \frac{1}{4} = \frac{9}{16}$.$h^2 + k^2 - 3h + k + \frac{10}{4} - \frac{9}{16} = 0$.$h^2 + k^2 - 3h + k + \frac{31}{16} = 0$.Multiplying by $16$,we get $16(h^2 + k^2) - 48h + 16k + 31 = 0$.Replacing $(h, k)$ with $(x, y)$,the locus is $16(x^2 + y^2) - 48x + 16y + 31 = 0$.

The equation of the locus of the midpoints of the chords of the circle $4x^2 + 4y^2 - 12x + 4y + 1 = 0$ that subtend an angle of $\frac{2\pi}{3}$ at its center is

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