A) \[\pm \frac{4}{3}\]
B) \[1\]
C) \[\frac{4}{3}\]
D) \[-\frac{4}{3}\]
Correct Answer: A
Solution :
Centres and radii of the given circles \[{{x}^{2}}+{{y}^{2}}=9\] and \[{{x}^{2}}+{{y}^{2}}+2a\alpha +2y+1=0\] is \[{{C}_{1}}(0,0),\,{{r}_{1}}=3\] and \[{{C}_{2}}(-\alpha ,1)\] and \[{{r}_{2}}=\sqrt{{{\alpha }^{2}}+1-1}=|\alpha |\] Since, two circles touch internally, \[\therefore \] \[{{C}_{1}}\,{{C}_{2}}={{r}_{1}}-{{r}_{2}}\] \[\Rightarrow \] \[\sqrt{{{\alpha }^{2}}+{{1}^{2}}}=3-|\alpha |\] \[\Rightarrow \] \[{{\alpha }^{2}}+1=9+{{\alpha }^{2}}-6|\alpha |\] \[\Rightarrow \] \[6|\alpha |=8\Rightarrow |\alpha |=\frac{4}{3}\] \[\Rightarrow \] \[\alpha =\pm \frac{4}{3}\]
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