Question

Find the locus of the middle points of the chords of the circle $x^2+y^2-2x-6y-10=0$ which pass through the origin.

Answer 1

$1st$ Method : The centre of circle is $(1,3)$ and $L(h,k)$ is the mid-point of the chord which passes through origin $(0,0)$.
Clearly $CL\perp AB\therefore m_1{m}_2=-1$.
$\therefore \frac{k-3}{h-1}\cdot \frac{k}{h}=-1$
or $h(h-1)+k(k-3)=0$.
Hence the locus of the mid-point $(h,k)$, is
$x^2+y^2-x-3y=0$.
Alt. Method : by $(T=S_1)$
If $(h,k)$ be the mid-point, then equation of chord by rule $T=S_1$ is
$x.h+y.k-(x+h)-3(y+k)-10=h^2+k^2-2h-6k-10$
It passes through origin.
$\therefore -h-3k-10=h^2+k^2-2h-6k-10$
Locus of mid-point $(h,k)$ is $x^2+y^2-x-3y=0$.