Question

If the lines $l_{1}x+m_{1}y+n_1=0$ and $1_{2}x+m_{2}y+n_2=0$ are conjugate lines with respect the circle $x^2+y^2+2gx+2fy+c=0$, then $l_1{l}_2+m_1{m}_2=$

• A. $(g{l}_1+f{m}_1+n_1)(g{l}_2+f{m}_2+n_2)$
• B. $\frac{n_1{n}_2}{g^2+f^2-c}$
• C. $\frac{n_1{n}_2}{\sqrt{g^2+f^2-c}}$
• D. $\frac{(g{l}_1+f{m}_1-n_1)(g{l}_2+f{m}_2-n_2)}{g^2+f^2-c}$

Answer 1

The correct option is D $\frac{(g{l}_1+f{m}_1-n_1)(g{l}_2+f{m}_2-n_2)}{g^2+f^2-c}$

$l_{1}x+m_{1}y+n_1=0$ has pole $P(h_1,k_1)$.
then $h_{1}n+k_{1}y+g(n+n_1)+f(y+k_1)+c=0$ -(1)
and $l_2{h}_1+m_2{k}_1+n_2=0$ -(2)
and $l_{2}n+m_{2}y+n_2=0$ has pole $(h_2,k_2)$
So,$x{h}_2+y{k}_2+g(n+h_2+f(y+k_2+c\left)\right)=0$ -(3)
$l_1{h}_2+m_1{k}_2+n_1=0$ -(4)
from (1),(2),(3)&(4)
$l_1{l}_2+m_1{m}_2=\frac{(g{l}_1+f{m}_1-n_1)\left(\right(g{l}_1+f{m}_2-n_2)}{g^2+f^2-c}$