Circle And System Of Circles Question 284
Question: If the circle $ x^{2}+y^{2}+6x-2y+k=0 $ bisects the circumference of the circle $ x^{2}+y^{2}+2x-6y-15=0, $ then k =
[EAMCET 2003]
Options:
A) 21
B) - 21
C) 23
D) - 23
Show Answer
Answer:
Correct Answer: D
Solution:
$ 2g_2(g_1-g_2),+2f_2(f_1-f_2)=c_1-c_2 $
$ 2(1)(3-1)+2(-3)(-1+3)=k+15 $
$ 4-12=k+15 $ or $ -8=k+15,\Rightarrow ,k=-23 $ .
BETA
