Question

The locus of the mid-points of the chords of the circle $x^2+y^2-2x-4y-11=0$ which subtend ${60}^o$ at the centre is

• A. $x^2+y^2-4x-2y-7=0$
• B. $x^2+y^2+4x+2y-7=0$
• C. $x^2+y^2-2x-4y-7=0$
• D. $x^2+y^2+2x+4y+7=0$

Answer 1

Answer: (C)

$x^2+y^2-2x-4y-7=0$

Let $AB$ be the chord of the circle and $P$ be the midpoint of $AB$.
It is known that perpendicular from the center bisects a chord.

Thus $△ACP$ is a right-angled triangle.

Now $AC=BC=$ radius.

The equation of the give circle can be written as
$(x-1{)}^2+(y-2{)}^2=16$

Hence, centre $C=(1,2)$ and radius $=r=4$ units.

$PC=AC\sin {60}^0$

$=r\sin {60}^0$

$=4\left(\frac{\sqrt{3}}{2}\right)$

$=2\sqrt{3}$ units

Therefore, $PC=2\sqrt{3}$

$\Rightarrow P{C}^2=12$

$\Rightarrow (x-1{)}^2+(y-2{)}^2=12$

$\Rightarrow x^2+y^2-2x-4y+5=12$

$\Rightarrow x^2+y^2-2x-4y-7=0$