The locus of the midpoints of the chords of t | vedclass

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(B) The given circle is $x^2 + y^2 + 4x - 6y - 12 = 0$.The center of the circle is $C(-2, 3)$ and the radius $R = \sqrt{(-2)^2 + 3^2 - (-12)} = \sqrt{

Vedclass · Accepted Answer

$(x + 2)^2 + (y - 3)^2 = 6.25$

Vedclass · Answer

(B) The given circle is $x^2 + y^2 + 4x - 6y - 12 = 0$.The center of the circle is $C(-2, 3)$ and the radius $R = \sqrt{(-2)^2 + 3^2 - (-12)} = \sqrt{4 + 9 + 12} = \sqrt{25} = 5$.Let the chord subtend an angle of $	heta = \frac{\pi}{3} = 60^\circ$ at the circumference.By the property of circles,the angle subtended by the chord at the center is $2	heta = 2 	imes 60^\circ = 120^\circ$.Let $(h, k)$ be the midpoint of the chord. The distance $p$ from the center $C(-2, 3)$ to the chord is given by $p = R \cos(\frac{2	heta}{2}) = R \cos(60^\circ)$.$p = 5 	imes \frac{1}{2} = 2.5$.The distance from the center $(-2, 3)$ to the point $(h, k)$ is $p$,so the equation is $(h - (-2))^2 + (k - 3)^2 = p^2$.$(h + 2)^2 + (k - 3)^2 = (2.5)^2 = 6.25$.Replacing $(h, k)$ with $(x, y)$,the locus is $(x + 2)^2 + (y - 3)^2 = 6.25$.

The locus of the midpoints of the chords of the circle $x^2 + y^2 + 4x - 6y - 12 = 0$ which subtend an angle of $\frac{\pi}{3}$ radians at its circumference is:

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