Question

Through a fixed point $O$ are drawn two straight lines $OPQ$ and $ORS$ to meet the circle in $P$ and $Q$, and $R$ and $S$, respectively. Prove that the locus of the point of intersection of $PS$ and $QR$, as also that of the point of intersection of $PR$ and $QS$, is the polar of $O$ with respect to the circle.

Answer 1

Let the two lines $OPQ$ and $ORS$ be taken as axes of $x$ and $y$ respectively, $0$ as origin and $\angle ROP=\omega$
Equation of the given circle maybe taken as
$x^2+y^2+2xy\cos \omega +2gx+2fy+c=\mathrm{0.....1}$
Now equation of x axis is $y=\mathrm{0.....2}$
Putting the value of y from $2$ in $1$, we get
$x^2+2gx+c=\mathrm{0....3}$
Let the co ordinates of P and Q be $(x_1,0)$ and $(x_2,0)$,$x_1$ and $x_2$ are given by; so
$x_1+x_2=-2g......4$ and $x_1{x}_2=c.....5$
Similarly if the co ordinates of R and S be $(0,y_1)$ and $(0,y_2)$
$y_1+y_2=2f.....6$
$y_1{y}_2=c.....7$
Equation of PS will be $\frac{x}{x_1}+\frac{y}{y_1}=\mathrm{1.....8}$
Equation of QR will be $\frac{x}{x_2}+\frac{y}{y_2}=\mathrm{1......9}$
Adding $8$ and $9$;-
$x\left(\frac{x_1+x_2}{x_1{x}_2}\right)+y\left(\frac{y_1+y_2}{y_1{y}_2}\right)=2$
$x\left(\frac{-2g}{c}\right)+y\left(\frac{-2f}{c}\right)=2$
or $gx+fy+c=0$
Polar of $(0,0)$ w.r.t. $1$ is
$x.0+y.0+g(x+0)+f(y+0)+c=0$
or $gx+fy+c=0$ which is same as $10$
Similarly we can prove that locus of the point of intersection of $PR$ and $QS$ is the same.