Circle And System Of Circles Question 364
Question: If a circle passes through the point (1, 2) and cuts the circle $ x^{2}+y^{2}=4 $ orthogonally, then the equation of the locus of its centre is
[MNR 1992]
Options:
A) $ x^{2}+y^{2}-3x-8y+1=0 $
B) $ x^{2}+y^{2}-2x-6y-7=0 $
C) $ 2x+4y-9=0 $
D) $ 2x+4y-1=0 $
Show Answer
Answer:
Correct Answer: C
Solution:
Let equation of circle be $ x^{2}+y^{2}+2gx+2fy+c=0 $ It passes through (1, 2) i.e. $ 1+4+2g+4f+c=0 $ and is orthogonal w.r.t $ x^{2}+y^{2}=4 $ . Therefore, $ 2g\times 0+2f\times 0=-c+4 $ or $ c=4 $
Hence, we get $ 2g+4f+9=0 $ , then $ 2x+4y-9=0 $ is the required locus.
BETA
