A) \[3\]
B) \[4\]
C) \[2\]
D) \[1\]
Correct Answer: B
Solution :
Given, equation of circles are \[{{x}^{2}}+\text{ }{{y}^{2}}=4,\]whose radius \[=2,\] centre \[=(0,0)\] and \[{{x}^{2}}+{{y}^{2}}-6x-8y-24=0,\] whose radius \[=\sqrt{9+16-24}=\sqrt{1}=1\] and centre \[=(3,4)\] Now, \[{{c}_{1}}\,{{c}_{2}}=\sqrt{{{(3-0)}^{2}}+{{(4-0)}^{2}}}\] \[=\sqrt{9+16}=5\] \[{{a}_{1}}+{{a}_{2}}=2+1=3\] Since, \[{{c}_{1}}{{c}_{2}}>{{a}_{1}}+{{a}_{2}}\] Number of common tangents \[=4\]
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