A) \[{{x}^{2}}+{{y}^{2}}-3x-8y+1=0\]
B) \[{{x}^{2}}+{{y}^{2}}-2x-6y-7=0\]
C) \[2x+4y-9=0\]
D) \[2x+4y-1=0\]
Correct Answer: C
Solution :
Let the equation of the circle be \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\] Since, this passes through\[(1,\,\,2)\]. \[\therefore \] \[{{1}^{2}}+{{2}^{2}}+2g(1)+2f(2)+c=0\] \[\Rightarrow \] \[5+2g+4f+c=0\] ... (i) Also, the circle\[{{x}^{2}}+{{y}^{2}}=4\]intersects the circle\[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\] orthogonally. \[\therefore \] \[2(g\cdot 0+f\cdot 0)=c-4\] \[\Rightarrow \] \[c=4\] On putting the value of c in Eq. (i), we get \[2g+4f+9=0\] Hence, the locus of centre\[(-g,\,\,-f)\]is \[-2x-4y+9=0\] or \[2x+4y-9=0\]
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