Question

The condition for the lines $lx+my+n=0$ and $l_{1}x+m_{1}y+n_1=0$ to be conjugate with respect to the circle $x^2+y^2=r^2$ is

• A. $r^2(l{l}_1+m{m}_1)=n{n}_1$
• B. $r^2(l{l}_1-m{m}_1)=n{n}_1$
• C. $r^2(l{l}_1+m{m}_1)+n{n}_1=0$
• D. $r^2(l{m}_1+l_{1}m)=n{n}_1$

Answer 1

Answer: (A)

$r^2(l{l}_1+m{m}_1)=n{n}_1$

The two lines are conjugate if pole of a point on the line lies on the other line.

Here, a point lying on the line $lx+my+n=0$ is $(x_1,\frac{-n-l{x}_1}{m})$

The equation of polar for the given circle is obtained as $x{x}_1+y{y}_1-r^2=0$

If substituted $(x_1,\frac{-n-l{x}_1}{m})$ into that line, we get $x{x}_1-\frac{y(n+l{x}_1)}{m}-r^2=0$, i.e. $mx{x}_1-y(n+l{x}_1)-m{r}^2=0$

Comparing this equation of the line with $l_{1}x+m_{1}y+n_1=0$, we get

$\frac{l_1}{m{x}_1}=\frac{m_1}{-(n+l{x}_1)}=\frac{n_1}{-m{r}^2}$

$\Rightarrow -m{l}_1{r}^2=m{n}_1{x}_1$

$\Rightarrow x_1=\frac{-l_1{r}^2}{n_1}$

and $m{m}_1{r}^2=n{n}_1+l{n}_1{x}_1$

$\Rightarrow m{m}_1{r}^2=n{n}_1+l{n}_1\times \frac{-l_1{r}^2}{n_1}$

$\Rightarrow r^2(m{m}_1+l{l}_1)=n{n}_1$