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  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2003

  • question_answer
    The number of common tangents to two circles \[{{x}^{2}}+{{y}^{2}}=4\] and\[{{x}^{2}}+{{y}^{2}}-8x+12=0\]is:

    A)  1                                            

    B)  2                            

    C)  5                            

    D)         4

    E)  3

    Correct Answer: E

    Solution :

    \[\because \]Centre and radius of circle\[{{x}^{2}}+{{y}^{2}}=4\]are \[{{C}_{1}}(0,0)\]and 2 respectively and centre and radius of circle\[{{x}^{2}}+{{y}^{2}}-8x+12=0\]are \[{{C}_{2}}(4,0)\]and \[2\]respectively. \[\because \]Here,                         \[{{C}_{1}}{{C}_{2}}={{r}_{1}}+{{r}_{2}}\] \[\therefore \]Circles touch each other externally so the number of common tangents is 3.


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