A) \[{{x}^{2}}+{{y}^{2}}+16x+12y+2=0\]
B) \[{{x}^{2}}+{{y}^{2}}-16x-12y-2=0\]
C) \[{{x}^{2}}+{{y}^{2}}-16x+12y+2=0\]
D) None of these
Correct Answer: C
Solution :
Let equation of circle be\[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\]. As it intersects orthogonally the given circles, we have \[2g+4f=6+c\] and \[4g+6f=2+c\]. As it passes through (1, 1), we have \[2g+2f=-2-c\] From these, we get \[g,\ f\] and c as ?8, 6, 2 respectively and hence equation of circle as \[{{x}^{2}}+{{y}^{2}}-16x+12y+2=0\].
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