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) ${r^2}\left( {l{l_1} + m{m_1}} \right) = n{n_1}$
(B) ${r^2}\left( {l{l_1} - m{m_1}} \right) = n{n_1}$
(C) ${r^2}\left( {l{l_1} + m{m_1}} \right) + n{n_1} = 0$
(D) ${r^2}\left( {l{m_1} + {l_1}m} \right) = n{n_1}$

Answer 1

Use the concept of conjugate lines which states that two lines are conjugate if the pole of a point on the line lies on the other line. Find the coordinate of any one point lying on the line $lx + my + n = 0$, then substitute the obtained point in the equation of polar for the given circle and then equate this equation to the given equation of line ${l_1}x + {m_1}y + {n_1} = 0$.

Complete step-by-step solution:
Let’s take a point $P\left( {{x_1},{y_1}} \right)$ on line $lx + my + n = 0$
Putting ${x_1}$ in the equation of line to find the corresponding $y$ coordinate, we get
$\Rightarrow l{x_1} + m{y_1} + n = 0$
On simplifying, we get
$\Rightarrow {y_1} = \dfrac{{ - l{x_1} - n}}{m}$
Therefore, the coordinate of point $P$ is $\left( {{x_1},\dfrac{{ - l{x_1} - n}}{m}} \right)$.
Polar of a point $P\left( {{x_1},{y_1}} \right)$ with respect to a circle ${x^2} + {y^2} = {r^2}$ is given by $x{x_1} + y{y_1} - {r^2} = 0 - - - (1)$.
Substituting point $P\left( {{x_1},{y_1}} \right)$ in $(1)$ , we get
$\Rightarrow x{x_1} + y\left( {\dfrac{{ - l{x_1} - n}}{m}} \right) - {r^2} = 0$
On solving,
$\Rightarrow mx{x_1} + y\left( { - l{x_1}y - ny} \right) - m{r^2} = 0$
$\Rightarrow mx{x_1} - y\left( {l{x_1}y + ny} \right) - m{r^2} = 0$
On rearranging,
$\Rightarrow \left( {m{x_1}} \right)x - \left( {l{x_1}y + ny} \right)y - m{r^2} = 0$
Since, the lines $lx + my + n = 0$ and ${l_1}x + {m_1}y + {n_1} = 0$ are conjugate so on comparing $\left( {m{x_1}} \right)x - \left( {l{x_1}y + ny} \right)y - m{r^2} = 0$ with the given equation of line ${l_1}x + {m_1}y + {n_1} = 0$, we get
$\Rightarrow \dfrac{{{l_1}}}{{m{x_1}}} = \dfrac{{{m_1}}}{{ - \left( {l{x_1} + n} \right)}} = \dfrac{{{n_1}}}{{ - \left( {m{r^2}} \right)}} - - - (2)$
On taking two at a time and solving we get,
$\Rightarrow \dfrac{{{l_1}}}{{m{x_1}}} = \dfrac{{{n_1}}}{{ - \left( {m{r^2}} \right)}}$
Taking $( - m{r^2})$ to left hand side and $m{x_1}$ to right hand side,
$\Rightarrow - m{r^2}{l_1} = m{n_1}{x_1}$
On simplifying,
$\Rightarrow {x_1} = \dfrac{{ - m{r^2}{l_1}}}{{m{n_1}}}$
$\Rightarrow {x_1} = \dfrac{{ - {r^2}{l_1}}}{{{n_1}}} - - - (3)$
Now taking second and third terms of $(2)$ , we get
$\Rightarrow \dfrac{{{m_1}}}{{ - \left( {l{x_1} + n} \right)}} = \dfrac{{{n_1}}}{{ - \left( {m{r^2}} \right)}}$
Taking $( - m{r^2})$ to the left-hand side and $- \left( {l{x_1} + n} \right)$ to the right-hand side, we get
$\Rightarrow - {m_1}m{r^2} = - {n_1}\left( {l{x_1} + n} \right)$
Eliminating minus sign from both the sides and on simplification, we get
$\Rightarrow {m_1}m{r^2} = {n_1}l{x_1} + {n_1}n$
Putting the value of ${x_1}$ from $(3)$ , we get
$\Rightarrow {m_1}m{r^2} = {n_1}l \times \left( {\dfrac{{ - {r^2}{l_1}}}{{{n_1}}}} \right) + {n_1}n$
On simplification,
$\Rightarrow {m_1}m{r^2} = - {r^2}l{l_1} + {n_1}n$
Taking $\left( { - {r^2}l{l_1}} \right)$ to the left-hand side,
$\Rightarrow {m_1}m{r^2} + {r^2}l{l_1} = {n_1}n$
Taking ${r^2}$ common from the left-hand side, we get
$\Rightarrow {r^2}\left( {{m_1}m + l{l_1}} \right) = {n_1}n$
On rewriting,
$\Rightarrow {r^2}\left( {l{l_1} + m{m_1}} \right) = n{n_1}$
Therefore, 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 ${r^2}\left( {l{l_1} + m{m_1}} \right) = n{n_1}$.
Hence, option (A) is correct.

Note: We have used that the polar of a point $P\left( {{x_1},{y_1}} \right)$ with respect to a circle ${x^2} + {y^2} = {r^2}$ is $x{x_1} + y{y_1} - {r^2} = 0$. But if the equation of circle is ${x^2} + {y^2} + 2gx + 2fy + c = 0$ then the equation of polar will be given by $x{x_1} + y{y_1} + g\left( {x + {x_1}} \right) + f\left( {y + {y_1}} \right) + c = 0$.