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The locus of the mid points of the chords of the circle $x^2+y^2+4 x-6 y-12=0$ which subtend an angle of $\pi$ /3 radians at its circumference is
Option: 1
$(x+2)^2+(y-3)^2=6.25$

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

Option: 3
$(x+2)^2+(y-3)^2=18.75$

Option: 4
$(x+2)^2+(y+3)^2=18.75$

Answers (1)

best_answer

Given

$\begin{gathered} \because \angle A C B=\pi / 3 \\ \therefore \angle A O B=\frac{2 \pi}{3} \end{gathered}$

$\Rightarrow \angle A O M=\angle B O M=\frac{\pi}{3}$

Let mid point of chord $AB\: is \left(x_1, y_1\right)\\$

$\therefore In \triangle A O M,$

$\cos \frac{\pi}{3}=\frac{O M}{O A}=\frac{\sqrt{\left(x_1+2\right)^2+\left(y_1-3\right)^2}}{5}$

$\text { or } \quad\left(\frac{5}{2}\right)^2=\left(x_1+2\right)^2+\left(y_1-3\right)^2$
$\Rightarrow \quad \frac{25}{4}=x_1^2+y_1^2+4 x_1-6 y_1+13$

$4 x_1^2+4 y_1^2+16 x_1-24 y_1+27=0$

$\therefore \text { Locus of Mid point is }$

$x^2+y^2+4 x-6 y+\frac{27}{4}=0$
$\Rightarrow \quad(x+2)^2+(y-3)^2=6.25$

Posted by

Ritika Jonwal

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