Circle And System Of Circles Question 303
Question: The value of k so that $ x^{2}+y^{2}+kx+4y+2=0 $ and $ 2(x^{2}+y^{2})-4x-3y+k=0 $ cut orthogonally is
[Karnataka CET 2004]
Options:
A) $ \frac{10}{3} $
B) $ \frac{-8}{3} $
C) $ \frac{-10}{3} $
D) $ \frac{8}{3} $
Show Answer
Answer:
Correct Answer: C
Solution:
Here, $ g_1=\frac{k}{2},,f_1=2,,c_1=2 $
$ g_2=-1,,f_2=\frac{-3}{4},,c_2=\frac{k}{2} $ Condition for orthogonal intersection,
therefore $ 2(g_1g_2+f_1f_2)=c_1+c_2 $
therefore $ 2,[ \frac{-k}{2}+( \frac{-3}{2} ) ]=2+\frac{k}{2} $
therefore $ -k-3=2+\frac{k}{2} $
therefore $ \frac{3k}{2}=-5 $ ; $ k=\frac{-10}{3} $ .
BETA
