Circle And System Of Circles Question 302
Question: The value of $ \lambda $ , for which the circle $ x^{2}+y^{2}+2\lambda x+6y+1=0 $ , intersects the circle $ x^{2}+y^{2}+4x+2y=0 $ orthogonally is
[MP PET 2004]
Options:
A) $ \frac{-5}{2} $
B) $ -1 $
C) $ \frac{-11}{8} $
D) $ \frac{-5}{4} $
Show Answer
Answer:
Correct Answer: D
Solution:
If two circles $ x^{2}+y^{2}+2g_1x+2f_1y+c_1=0 $ and $ 2(g_1g_2+f_1f_2)=c_1+c_2 $ intersect orthogonally then they must follow $ 2(g_1g_2+f_1f_2)=c_1+c_2 $ and $ g_1=\lambda ,,f_1=3,,c_1=1 $ and $ g_2=2,,f_2=1,,c_2=0 $
So, $ 2,(2\lambda +3)=1+0\Rightarrow 2\lambda +3=\frac{1}{2} $
$ \Rightarrow \lambda =\frac{-5}{4} $ .
BETA
