A) 0
B) -1
C) \[\frac{1}{2}\]
D) \[-\frac{1}{2}\]
Correct Answer: D
Solution :
[d]: Let the equation of the circle be \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0.\] \[\because \]It cuts the two given circles orthogonally \[\therefore \]\[-8g-2f=c+16\]and\[-4g-4f=c+1\] \[\Rightarrow \]\[8g+2f+c=-16\] ?(i) and\[4g+4f+c=-1\] \[\because \]The circle passes through (1,1) \[\therefore \]\[2g+2f+c=-2\] ...(iii) Solving (i), (ii) and (iii) we get \[g=-\frac{7}{3},f=\frac{17}{6},c=-3\] \[\therefore \]The centre is \[\left( \frac{7}{3},-\frac{17}{6} \right)=(a,b)\] \[\Rightarrow \]\[a=\frac{7}{3}\]and\[b=-\frac{17}{6}\Rightarrow a+b=-\frac{1}{2}\]
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