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JEE Main & Advanced AIEEE Solved Paper-2002

  • question_answer
    The greatest distance of the point P(10,7) from the circle \[{{x}^{2}}+{{y}^{2}}-4x-2y-20=0\] is   AIEEE  Solved  Paper-2002

    A) 10 units       

    B)           15 units

    C)           5 units                     

    D)           None of these

    Correct Answer: B

    Solution :

       Equation of circle is                    \[{{x}^{2}}+{{y}^{2}}-4x-2y-20=0\]. whose centre is C(2,1) and radius is 5 units. At point? (10,7), \[{{S}_{1}}={{10}^{2}}+{{7}^{2}}-4\times 10-2\times 7-20>0\] So, P lies outside the circle. Now, \[PC=\sqrt{{{(2-10)}^{2}}+{{(1-7)}^{2}}}\]                 \[=\sqrt{{{8}^{2}}+{{6}^{2}}}=\sqrt{{{10}^{2}}}=10\] \[\therefore \] Greatest distance between circle and the point \[p=10+5=15\] units.                     


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