JEE Main & Advanced AIEEE Solved Paper-2004

  • question_answer
    A variable circle passes through the fixed point \[A(p,\text{ }q)\]and touches x-axis. The locus of the other end of the diameter through A is

    A) \[{{(x-p)}^{2}}=4qy\]     

    B) \[{{(x-q)}^{2}}=4qy\]

    C)                        \[{{(y-p)}^{2}}=4qx\]     

    D) \[{{(y-q)}^{2}}=4px\]

    Correct Answer: A

    Solution :

    If\[A({{x}_{1}},{{y}_{1}})\]and\[B({{x}_{2}},{{y}_{2}})\]are the coordinates of end points diameter of a circle, then the equation of circle is \[({{x}_{1}},{{y}_{1}})(x-{{x}_{2}})+(y-{{y}_{1}})(y-{{y}_{2}})=0\] In a circle AB is as a diameter where the coordinates of A are\[(p,\text{ }q)\]and let the coordinates of \[({{x}_{1}},{{y}_{1}})\]. Equation of circle in diameter form is \[(x-p)(x-{{x}_{1}})+(y-p)(y-{{y}_{1}})=0\] \[\Rightarrow \]\[{{x}^{2}}-(p+{{x}_{1}})x+p{{x}_{1}}+{{y}^{2}}\]                                 \[-({{y}_{1}}+q)y+q{{y}_{1}}=0\] \[\Rightarrow \]\[{{x}^{2}}-(p+{{x}_{1}})x+{{y}^{2}}\] \[-({{y}_{1}}+q)y+p{{x}_{1}}+q{{y}_{1}}=0\] Since, this circle touches X-axis. \[\therefore \]  \[y=0\] \[\Rightarrow \] \[{{x}^{2}}-(p+{{x}_{1}})x+p{{x}_{1}}+q{{y}_{1}}=0\] Also, the discriminant of above equation will be equal to zero because circle touches X-axis. \[\therefore \]  \[{{(p+{{x}_{1}})}^{2}}=4(p{{x}_{1}}+q{{y}_{1}})\] \[\Rightarrow \]               \[{{p}^{2}}+x_{1}^{2}+2p{{x}_{1}}=4p{{x}_{1}}+4q{{y}_{1}}\] \[\Rightarrow \]               \[x_{1}^{2}-2p{{x}_{1}}+{{p}^{2}}=4q{{y}_{1}}\] Hence, the locus of point S is \[{{(x-p)}^{2}}=4qy\]

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