Question

Find the locus of the mid-points of the chords of the circle $x^2+y^2=a^2$ which subtend a right angle at any point $(\alpha ,\beta )$.

Answer 1

$x2+y^2=a^2......\left(1\right)$
$hx+ky=h^2+k^2.....\left(2\right)$
where $(h,k)$ is middle point of chord $AB$, by $T=S_1$.
$AP$ and $PB$ are perpendiculars as $AB$ subtends an angle of ${90}^{\circ }$ at $P(\alpha ,\beta )$.
If $A$ and $B$ be $(x_1,y_1)$ and $(x_2,y_2)$ respectively $\frac{y_1-\beta }{x_1-\alpha }\cdot \frac{y_2-\beta }{x_2-\alpha }=-1$
or $(x_1-\alpha )(x_1-\alpha )+(y_1-\beta )(y_2-\beta )=0$
or $(x_1{x}_2+y_1{y}_2)-\alpha (x_1+x_2)-\beta (y_1+y_2)+{\alpha }^2+{\beta }^2=0.....\left(3\right)$
Eliminating $y$ between $\left(1\right)$ and $\left(2\right)$, we will get a quadratic
$x^2(h^2+k^2)-2\lambda hx+{\lambda }^2-a^2{k}^2=0$
$\lambda x^2-2\lambda hx+{\lambda }^2-a^2{k}^2=0$
Similarly $\lambda y^2-2\lambda ky+{\lambda }^2-a^2{h}^2=0$
Putting the values of $x_1{x}_2,y_1{y}_2,x_1+x_2,y_1+y_2$ from above in $\left(3\right)$, we get
$\frac{{\lambda }^2-a^2{k}^2+{\lambda }^2-a^2{h}^2}{\lambda }-\alpha .2h-\beta .2k+{\alpha }^2+{\beta }^2=0$
or $2(\lambda -h\alpha -k\beta )+({\alpha }^2+{\beta }^2)-a^2=0$
Dividing by $2$ and putting $\lambda =h^2+k^2$ and then generalizing
$x^2+y^2-\alpha x-\beta y+\frac{1}{2}({\alpha }^2+{\beta }^2-a^2)=0$.