Circle And System Of Circles Question 274
Question: The locus of the centre of circle which cuts the circles $ x^{2}+y^{2}+4x-6y+9=0 $ and $ x^{2}+y^{2}-4x+6y+4=0 $ orthogonally is
[UPSEAT 2001]
Options:
A) $ 12x+8y+5=0 $
B) $ 8x+12y+5=0 $
C) $ 8x-12y+5=0 $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Let the given circle be $ x^{2}+y^{2}+2hx+2ky+c=0 $ Since the circle cuts $ x^{2}+y^{2}+4x-6y+9=0 $ and $ x^{2}+y^{2}-4x+6y+4=0 $ orthogonally, we have $ 2h(2)+2k(-3)=c+9 $
$ \Rightarrow $ $ 4h-6k=c+9 $ …..(i) and $ 2h(-2)+2k(3),=c+4 $
$ \Rightarrow $ $ -4h+6k=c+4 $ …..(ii) From (i) and (ii); $ c+9=-,c-4 $
$ \Rightarrow $ $ 2c=-13 $ …..(iii) From (i), $ 8h-12k=2c+18 $
$ \Rightarrow $ $ 8h-12k=5 $ …..(iv) Centre of the given circle is $ (-,h,-,k) $ .
Hence locus of $ (-h,,-k) $ from (iv) we have, $ 8(-x)-12(-y)=5 $
$ \Rightarrow $ $ 8x-12y+5=0. $
BETA
