A) \[1\]
B) \[2\]
C) \[3\]
D) \[4\]
Correct Answer: A
Solution :
Let\[{{S}_{1}}\equiv {{x}^{2}}+{{y}^{2}}-2x-1=0\] Centre,\[{{C}_{1}}=(1,\,\,0)\]and Radius \[{{r}_{1}}=\sqrt{1+0+1}=\sqrt{2}\] and \[{{S}_{2}}={{x}^{2}}+{{y}^{2}}-2y-7=0\] Centre,\[{{C}_{2}}(0,\,\,1)\] and Radius \[{{r}_{2}}=\sqrt{0+1+7}=2\sqrt{2}\] Here, \[{{C}_{1}}{{C}_{2}}=\sqrt{{{(1-0)}^{2}}+{{(0-1)}^{2}}}=\sqrt{2}\] \[\because \] \[{{C}_{1}}{{C}_{2}}={{r}_{2}}-{{r}_{1}}\] \[\therefore \]Circles touch each other internally and so only one common tangent can be drawn to given two circles.
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