A) \[12x+8y+5=0\]
B) \[8x+12y+5=0\]
C) \[8x-12y+5=0\]
D) None of these
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, \[8(-x)-12(-y)=5\] \[\Rightarrow \]\[8x-12y+5=0.\]
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