Circle And System Of Circles Question 261
Question: If the circles $ x^{2}+y^{2}=4,x^{2}+y^{2}-10x+\lambda =0 $ touch externally, then $ \lambda $ is equal to
[AMU 1999]
Options:
A) -16
B) 9
C) 16
D) 25
Show Answer
Answer:
Correct Answer: A
Solution:
Circles x2 + y2 = 4, x2 + y2 - 10x + l = 0 touch externally \ $ C_1C_2=r_1+r_2 $
therefore $ C_1(0,,0) $ and $ C_2=(5,,0) $
$ r_1=2 $ and $ r_2=\sqrt{25+\lambda } $ \ $ \sqrt{{{(5-0)}^{2}}+0}=2+\sqrt{25+\lambda } $
therefore $ 5-2=\sqrt{25+\lambda } $
therefore $ 3=\sqrt{25+\lambda } $
therefore $ 9=25+\lambda $
therefore l = - 16.
BETA
