Circle And System Of Circles Question 237
Question: The equation of circle which passes through the point (1,1) and intersect the given circles $ x^{2}+y^{2}+2x+4y+6=0 $ and $ x^{2}+y^{2}+4x+6y+2=0 $ orthogonally, is
Options:
A) $ x^{2}+y^{2}+16x+12y+2=0 $
B) $ x^{2}+y^{2}-16x-12y-2=0 $
C) $ x^{2}+y^{2}-16x+12y+2=0 $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Let equation of circle be $ x^{2}+y^{2}+2gx+2fy+c=0 $ .
As it intersects orthogonally the given circles, we have $ 2g+4f=6+c $ and $ 4g+6f=2+c $ .
As it passes through (1, 1), we have $ 2g+2f=-2-c $
From these, we get $ g,\ f $ and c as -8, 6, 2 respectively and
Hence equation of circle as $ x^{2}+y^{2}-16x+12y+2=0 $ .
BETA
