Question

The locus of the mid point of all chords of the circle $x^2+y^2-2x-2y=0$ such that the pair of lines joining $(0,0)\&$ the point of intersection of the chords with the circles make equal angle with axis of $x$, is

• A. $x-y=2$
• B. $x+y=2$
• C. $x+y=-2$
• D. None of these

Answer 1

Answer: (B)

Let mid-point $\equiv (h,k)$
$\therefore h^2+k^2-2h-2k=hx+ky-(x+h)-(y+k)$
$\Rightarrow h^2+k^2-h-k=hx+ky-x-y$
$\Rightarrow \frac{(h-1)x+(k-1)y}{h^2+k^2-h-k}=1$
Homogenization
$x^2+y^2-2x\left(\frac{(h-1)x+(k-1)y}{h^2+k^2-h-k}\right)$
$-2y\left(\frac{(h-1)x+(k-1)y}{h^2+k^2-h-k}\right)=0$
$\therefore \frac{-2(k-1)-2(h-1)}{h^2+k^2-h-k}=0$
$\Rightarrow -2k+2-2h+2=0$
$\Rightarrow h+k=2$
$\therefore$ required locus :
$x+y=2$