Question

Find the locus of the mid point of the chord of a circle $x^2+y^2=4$ such that the segment intercepted by the chord on the curve $x^2-2x-2y=0$ subtends a right angle at the origin.

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

Answer 1

Answer: (A)

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

Let the midpoint be $(h,k)$

Equation of chord $T=S$

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

Let the above line intersects curve $x^2-2x-2y=0$ at $A$ and $B$.

Let $O$ be the origin.

For finding $OA\times OB$, we will use homogenization.

$x^2-2x\left(\frac{(hx+ky)}{h^2+k^2}\right)-2y\left(\frac{hx+ky}{h^2+k^2}\right)=0$

$x^2(h^2+k^2)-x(hx+ky)-2y(hx+ky)=0$

$\Rightarrow x^2(h^2+k^2)+x^2(-2h)-2kxy-2hxy-2k{y}^2=0$

$\Rightarrow x^2(h^2+k^2-2h)+y^2(-2k)+y(-2hx-2kx)=0$

$\because OA$ and $OB$ are perpendicular.

$\Rightarrow h^2+k^2-2h-2k=0$

$\Rightarrow h^2+k^2=2(h+k)$

$\therefore$ Locus is

$x^2+y^2=2(x+y)$

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