Circle And System Of Circles Question 362
Question: If the lines $ l_1x+m_1y+n_1=0 $ and $ l_2x+m_2y+n_2=0 $ cuts the axes at con-cyclic points, then
Options:
A) $ l_1l_2=m_1m_2 $
B) $ l_1m_1=l_2m_2 $
C) $ l_1l_2+m_1m_2=0 $
D) $ l_1m_2=l_2m_1 $
Show Answer
Answer:
Correct Answer: A
Solution:
$ P_1\equiv ( -\frac{n_1}{l_1},\ 0 ) $ , $ P_2\equiv ( 0,\ \frac{-n_1}{m_1} ) $ , $ P_3\equiv ( -\frac{n_2}{l_2},\ 0 ) $ and $ P_4\equiv ( 0,\ -\frac{n_2}{m_2} ) $
$ {\angle P_1P_2P_3=\angle P_1P_4P_3} $ Now, $ m_{12}=-\frac{l_1}{m_1},\ m_{23}=-\frac{n_1}{n_2}.\frac{l_2}{m_1},\ m_{14}=-\frac{n_2}{n_1}.\frac{l_1}{m_2} $ , $ m_{34}=-\frac{l_2}{m_2} $
$ \tan \theta =\frac{-\frac{l_1}{m_1}+\frac{n_1l_2}{n_2m_1}}{1+\frac{n_1l_1l_2}{n_2m_1^{2}}} $ and $ \tan \varphi =\frac{-\frac{n_2l_1}{n_1m_2}+\frac{l_2}{m_2}}{1+\frac{n_2l_1l_2}{n_1m_2^{2}}} $ Now, $ \tan \theta =\tan \varphi \Rightarrow m_1m_2=l_1l_2 $ Aliter: Line $ l_1x+m_1y+n_1=0 $ cuts x and y-axes in $ A( -\frac{n_1}{l_1},\ 0 ) $ , $ B( 0,\ -\frac{n_1}{m_1} ) $ and line $ l_2x+m_2y+n_2=0 $ cuts axes in $ C( -\frac{n_2}{l_2},\ 0 ) $ , $ D( 0,\ \frac{-n_2}{m_2} ) $ . So AC and BD are chords along x and y-axes intersecting at origin O. Since A, B, C, D are concyclic, so OA.OC = OB.OD or $ | ( -\frac{n_1}{l_1} )( -\frac{n_2}{l_2} ) |=| ( -\frac{n_1}{m_1} )( -\frac{n_2}{m_2} ) | $ or $ |l_1l_2|\ =\ |m_1m_2| $ So $ l_1l_2=m_1m_2 $ is correct among the given choices, which is given in (a).
BETA
