Circle And System Of Circles Question 260
Question: The equation of a circle that intersects the circle $ x^{2}+y^{2}+14x+6y+2=0 $ orthogonally and whose centre is (0, 2) is
[MP PET 1998]
Options:
A) $ x^{2}+y^{2}-4y-6=0 $
B) $ x^{2}+y^{2}+4y-14=0 $
C) $ x^{2}+y^{2}+4y+14=0 $
D) $ x^{2}+y^{2}-4y-14=0 $
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Answer:
Correct Answer: D
Solution:
In circle, $ x^{2}+y^{2}+14x+6y+2=0 $
$ g=7,\ f=3,\ c=2 $
Centre of circle $ (-g,\ -f)=(0,\ 2) $ , (Given) For orthogonally intersection, $ 2gg’+2ff’=c+c’ $
$ 0-12=2+c’\Rightarrow c’=-14 $
Put the values, in equation $ x^{2}+y^{2}+2g’x+2f’x+c’=0 $ .
$ \Rightarrow x^{2}+y^{2}+0-4y-14=0\Rightarrow x^{2}+y^{2}-4y-14=0 $ .
BETA
