Circle And System Of Circles Question 211
Question: The equation of the circle which passes through the point of intersection of circles $ x^{2}+y^{2}-8x-2y+7=0 $ and $ x^{2}+y^{2}-4x+10y+8=0 $ and having its centre on $ y $ -axis, will be
Options:
A) $ x^{2}+y^{2}+22x+9=0 $
B) $ x^{2}+y^{2}+22x-9=0 $
C) $ x^{2}+y^{2}+22y+9=0 $
D) $ x^{2}+y^{2}+22y-9=0 $
Show Answer
Answer:
Correct Answer: C
Solution:
Using $ S_1+\lambda S_2=0 $ , but its centre is on y-axis. i.e., $ -8-4\lambda =0 $ or $ \lambda =-2 $ .
Hence required equation is $ x^{2}+y^{2}+22y+9=0 $ .
BETA
