JEE Main & Advanced JEE Main Paper (Held On 12-Jan-2019 Evening)

  • question_answer
    Let \[{{z}_{1}}\] and \[{{z}_{2}}\]be two complex numbers satisfying \[|{{z}_{1}}|=9\] and \[|{{z}_{2}}-3-4i|=4\].Then the minimum value of \[|{{z}_{1}}-{{z}_{2}}|\]is [JEE Main Online Paper Held On 12-Jan-2019 Evening]

    A) \[\sqrt{2}\]                                

    B) 2    

    C) 0                                 

    D)   1

    Correct Answer: C

    Solution :

    Given, \[|{{z}_{1}}|=9,\]represents a circle with centre \[{{C}_{1}}(0,0)\] and radius \[{{r}_{1}}=9\]. Also, \[|{{z}_{2}}-(3+4i)|=4\]represents a circle with centre \[{{C}_{2}}(3,4)\]and radius \[{{r}_{2}}=4.\] \[\therefore \]\[{{C}_{1}}{{C}_{2}}=|{{r}_{1}}-{{r}_{2}}|=5\] So, circles touches each other internally. \[\therefore \]  \[|{{z}_{1}}-{{z}_{2}}{{|}_{\min }}=0\]


You need to login to perform this action.
You will be redirected in 3 sec spinner