Question

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

• A. (x-2)^2 + (y + 3)^2 = 6.25
• B. (x+2)^2+(y-3)^2 = 6.25
• C. (x+2)^2 + (y-3)^2 = 18.75
• D. (x+2)^2 + (y + 3)^2 = 18.75

Answer 1

Answer: (B)

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

Answer 2

The equation of the given circle is $x^2+y^2+4x-6y-12=0$. The center of this circle is $C(-2, 3)$ and its radius is $R=5$. Let $AB$ be a chord of this circle that subtends an angle $\frac{\pi}{3}$ at the circumference. The angle subtended by this chord at the center $C$ is $2 \times \frac{\pi}{3} = \frac{2\pi}{3}$. Let $M$ be the midpoint of the chord $AB$. The distance $CM$ from the center to the midpoint of the chord is given by $CM = R \cos\left(\frac{\text{angle at center}}{2}\right)$. So, $CM = 5 \cos\left(\frac{2\pi/3}{2}\right) = 5 \cos\left(\frac{\pi}{3}\right) = 5 \times \frac{1}{2} = 2.5$. The locus of the midpoint $M$ is a circle with the same center $C(-2, 3)$ and a radius $r' = 2.5$. The equation of this locus is $(x - (-2))^2 + (y - 3)^2 = (2.5)^2$, which simplifies to $(x+2)^2 + (y-3)^2 = 6.25$.