Circle And System Of Circles Question 10
Question: The two circles which passes through $ (0,a) $ and $ (0,-a) $ and touch the line $ y=mx+c $ will intersect each other at right angle, if
Options:
A) $ a^{2}=c^{2}(2m+1) $
B) $ a^{2}=c^{2}(2+m^{2}) $
C) $ c^{2}=a^{2}(2+m^{2}) $
D) $ c^{2}=a^{2}(2m+1) $
Show Answer
Answer:
Correct Answer: C
Solution:
Equation of circles $ [x^{2}+(y-a)(y+a)]+\lambda x=0\Rightarrow x^{2}+y^{2}+\lambda x-a^{2}=0 $ and $ \sqrt{{{( \frac{\lambda }{2} )}^{2}}+a^{2}}=\frac{\frac{-m\lambda }{2}+c}{\sqrt{1+m^{2}}} $
$ \Rightarrow (1+m^{2})[ \frac{{{\lambda }^{2}}}{4}+a^{2} ]={{( \frac{m\lambda }{2}-c )}^{2}} $
$ \Rightarrow (1+m^{2})[ \frac{{{\lambda }^{2}}}{4}+a^{2} ]=\frac{m^{2}{{\lambda }^{2}}}{4}-mc\lambda +c^{2} $
$ \Rightarrow {{\lambda }^{2}}+4mc\lambda +4a^{2}(1+m^{2})-4c^{2}=0 $
$ \therefore $ $ {\lambda_1}{\lambda_2}=4[a^{2}(1+m^{2})-c^{2}]\Rightarrow g_1g_2=[a^{2}(1+m^{2})-c^{2}] $ and $ g_1g_2+f_1f_2=\frac{c_1+c_2}{2}\Rightarrow a^{2}(1+m^{2})-c^{2}=-a^{2} $
Hence $ c^{2}=a^{2}(2+m^{2}) $ .
BETA
