Question

The locus of the mid points of the chord of the circle $x^2+y^2=4$, which subtended a right angle at the origin is

• A. $x+y=1$
• B. $x^2+y^2=1$
• C. $x+y=2$
• D. $x^2+y^2=2$

Answer 1

Answer: (D)

$x^2+y^2=2$

Let $(h,k)$ be the mid-point of a chord of circle $x^2+y^2=4$.

Therefore, the equation of chord-

$hx+ky-4=h^2+k^2-4$

$\Rightarrow hx+ky=h^2+k^2.....\left(1\right)$

Equation of straight line joining the origin to the points of intersection of

$\left(1\right)$ and the circle $x^2+y^2-4$-

$x^2+y^2=4{\left(\frac{hx+ky}{h^2+k^2}\right)}^2.....\left(2\right)$

Since the above pair of lines subtends a the right angle.

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

Hence the locus of the mid-point of the chord is $x^2+y^2=2$.