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

Hint: 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 \[x{x_1} + y{y_1} + g\left( {x + {x_1}} \right) + f\left( {y + {y_1}} \right) + c = 0\].