A) \[1\]
B) \[2\]
C) \[5\]
D) \[3\]
Correct Answer: D
Solution :
Since, centre and radius of a circle \[{{x}^{2}}+{{y}^{2}}=4\] are \[{{C}_{1}}(0,\,\,0)\] and \[2\] respectively and centre and radius of another circle \[{{x}^{2}}+{{y}^{2}}-8x+12=0\] are \[{{C}_{2}}(4,\,\,0)\]and \[2\] respectively. Now, \[{{C}_{1}}{{C}_{2}}=\sqrt{{{(4-0)}^{2}}+0}=4\] and \[{{r}_{1}}+{{r}_{2}}=2+2=4\] \[\because \] \[{{C}_{1}}{{C}_{2}}={{r}_{1}}+{{r}_{2}}\] \[\therefore \]Two circles touch each other externally, so the number of common tangents is\[3\]. Note: (i) If\[{{C}_{1}}{{C}_{2}}={{r}_{1}}-{{r}_{2}}\], one tangents are possible. If\[{{C}_{1}}{{C}_{2}}>{{r}_{1}}+{{r}_{2}}\], four tangents are possible.
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