Circle And System Of Circles Question 197
Question: The equation of the circle passing through the point (-2, 4) and through the points of intersection of the circle $ x^{2}+y^{2}-2x-6y+6=0 $ and the line $ 3x+2y-5=0 $ , is
[RPET 1996]
Options:
A) $ x^{2}+y^{2}+2x-4y-4=0 $
B) $ x^{2}+y^{2}+4x-2y-4=0 $
C) $ x^{2}+y^{2}-3x-4y=0 $
D) $ x^{2}+y^{2}-4x-2y=0 $
Show Answer
Answer:
Correct Answer: B
Solution:
Required equation of the circle, $ (x^{2}+y^{2}-2x-6y+6)+\lambda (3x+2y-5)=0 $
This circle passing through points $ (-2,\ 4) $ , therefore $ (4+16+4-24+6)+\lambda (-6+8-5)=0 $ ,
$ \therefore \ \lambda =2 $
$ \therefore \ (x^{2}+y^{2}-2x-6y+6)+2(3x+2y-5)=0 $
$ \Rightarrow x^{2}+y^{2}+4x-2y-4=0 $ .
BETA
