Question

The equation of the locus of the mid-points of chord of the circle $4x^2+4y^2-12x+4y+1=0$ that subtends and angle of $\frac{2\pi }{3}$ at its centre is $x^2+y^2-kx+y+\frac{31}{16}=0$ then $k$ is__

Answer 1

Consider the equation of the circle.

$4x^2+4y^2-12x+4y+1=0$

$x^2+y^2-3x+y+\frac{1}{4}=0$

Therefore, centre of this circle is,

$C=(\frac{3}{2},-\frac{1}{2})$

Radius of this circle $=\sqrt{\frac{9}{4}+\frac{1}{4}-\frac{1}{4}}=\frac{3}{2}$

Consider the diagram shown below.

Here, chord $AB$ subtends an angle of $\frac{2\pi }{3}$ at the centre of circle. Let the mid-point of the chord $M(h,k)$. Therefore,

$\sin 30{}^{\circ }=\frac{CM}{AM}$

$\frac{1}{2}=\frac{\sqrt{{(h-\frac{3}{2})}^2+{(k+\frac{1}{2})}^2}}{\frac{3}{2}}$

${(h-\frac{3}{2})}^2+{(k+\frac{1}{2})}^2={\left(\frac{3}{4}\right)}^2$

$h^2+k^2-3h+k+\frac{9}{4}+\frac{1}{4}-\frac{9}{16}=0$

$h^2+k^2-3h+k+\frac{31}{16}=0$

Substitute $(x,y)$ for $(h,k)$.

$x^2+y^2-3x+y+\frac{31}{16}=0$

Comparing this with the given equation of the locus of the mid-point, we have

$\Rightarrow k=3$

Hence, this is the required result.