Question

The locus of the midpoints of the chords of the circle $4x^2+4y^2-12x+4y+1=0$ that subtend an angle of $\frac{2\pi }{3}$ at its centre is

• A. $x^2+y^2-3x+y-\frac{31}{16}=0$
• B. $x^2+y^2-3x+y+\frac{31}{16}=0$
• C. $x^2+y^2+3x+y+\frac{31}{16}=0$
• D. $x^2+y^2-3x-y+\frac{31}{16}=0$

Answer 1

The correct option is B $x^2+y^2-3x+y+\frac{31}{16}=0$

The circle $4x^2+4y^2-12x+4y+1=0$ can also be written as
$x^2+y^2-3x+y+\frac{1}{4}=0$
which is of the form $x^2+y^2+2gx+2fy+c=0$
Centre $C=(-g,-f)=(\frac{3}{2},-\frac{1}{2})$
radius$=\sqrt{g^2+f^2-c}=\sqrt{{\left(\frac{3}{2}\right)}^2+{\left(\frac{-1}{2}\right)}^2-\frac{1}{4}}=\frac{3}{2}$
$AB$ is a chord with midpoint $M(x,y)$
$CB=\frac{3}{2},\angle PCB=\frac{\pi }{3}$
$\therefore CP=CB.\cos \frac{\pi }{3}=\frac{3}{2}.\frac{1}{2}=\frac{3}{4}$
$P$ traces a circle with centre $C$ and radius $\frac{3}{4}$
$\therefore {(x-\frac{3}{2})}^2+{(y+\frac{1}{2})}^2={\left(\frac{3}{4}\right)}^2$
or $x^2+y^2-3x+y+\frac{9}{4}+\frac{1}{4}-\frac{9}{16}=0$
On simplification, we get
$x^2+y^2-3x+y+\frac{31}{16}=0$