Question

Let $C$ be the circle with centre $\left(0,0\right)$ and radius $3$ units. The equation of the locus of the mid-points of the chords of the circle $C$ that subtend an angles of $\frac{2\pi }{3}$ at its centre, is:

• A. $x^3+y^2=1$
• B. $x^2+y^2=\frac{27}{4}$
• C. $x^2+y^2=\frac{9}{4}$
• D. $x^2+y^2=\frac{3}{2}$

Answer 1

The correct option is C $x^2+y^2=\frac{9}{4}$

Let chord $AB$ subtends an angle $\frac{2\pi }{3}$ at the center.

Let $OP$ be the perpendicular bisector of the chord $AB$ as shown in figure.

$OA=OB=r$ [ Given ]

$\angle AOB=\frac{2\pi }{3}$

Since, $OP$ is the perpendicular bisector so

$\angle AOP=\angle POB=\theta$

$\Rightarrow$ $\theta =\frac{\pi }{3}$

The equation of the circle is $x^2+y^2=9.$

So, the equation of the chord with mid-point at $P(h,k)$ will be,

$\Rightarrow$ $hx+ky-9=h^2+k^2-9$

$\Rightarrow$ $hx+ky=h^2+k^2$ ----- ( 1 )

In $△OPB,$

$\cos \theta =\frac{OP}{OB}$, where $OP=\sqrt{h^2+k^2}$ and $OB=r$

$\Rightarrow$ $r\cos \theta =\sqrt{h^2+k^2}$

$\Rightarrow$ $3\times \cos \frac{\pi }{3}=\sqrt{h^2+k^2}$

$\Rightarrow$ $3\times \frac{1}{2}=\sqrt{h^2+k^2}$

$\Rightarrow$ $h^2+k^2=\frac{9}{4}$ ------ ( 2 )

From ( 1 ) and ( 2 ) the locus of the mid-point $(h,k)$ will be

$\Rightarrow$ $hx+ky=\frac{9}{4}$

Now replace, $h=x$ and $k=y$

$\Rightarrow$ $x^2+y^2=\frac{9}{4}$