A) \[a=5\] and \[b\notin (-1,1)\]
B) \[a=1\] and \[b\notin (-1,1)\]
C) \[a=2\] and \[b\notin (-\infty ,1)\]
D) \[a=5\] and \[b\in (-\infty ,1)\]
Correct Answer: D
Solution :
On solving the equation of diameters, we get \[x=1\] and y = 1 i.e., centre of circle is (1, 1) Given also \[{{x}^{2}}+{{y}^{2}}-2({{a}^{2}}-7a+11)x\] \[-2({{a}^{2}}-6a+6)y+{{b}^{3}}+1=0\] ...(i) Centre of the circle is \[\{{{a}^{2}}-7a+11,\,{{a}^{2}}-6a+6\}\] \[\therefore \] \[{{a}^{2}}-7a+11=1\] and \[{{a}^{2}}-6a+6=1\] \[\Rightarrow \] \[{{a}^{2}}-7a+10=0\] and \[{{a}^{2}}-6a+5=0\] \[\Rightarrow \] \[a=2,\,5\] and \[a=1,5\] Put, \[a=5\] in Eq. (i) we get \[{{x}^{2}}+{{y}^{2}}-2x-2y+{{b}^{3}}+1=0\] \[\Rightarrow \] \[{{(x-1)}^{2}}+{{(y-1)}^{3}}={{\left( \sqrt{1-{{b}^{3}}} \right)}^{2}}\] Here, Radius \[=\sqrt{1-{{b}^{3}}}>0\] \[\Rightarrow \] \[{{b}^{3}}-1<0\] \[\Rightarrow \] \[(b-1)({{b}^{2}}+1+b)<0\] \[\Rightarrow \] \[b<1\] i.e., \[b\in (-\infty ,\,1)\]
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