JEE Main & Advanced Mathematics Circle and System of Circles Question Bank Tangent and normal to a circle

  • question_answer
    If the line \[y=mx+c\]be a tangent to the circle \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\], then the point of contact is

    A)            \[\left( \frac{-{{a}^{2}}}{c},{{a}^{2}} \right)\]                  

    B)            \[\left( \frac{{{a}^{2}}}{c},\frac{-{{a}^{2}}m}{c} \right)\]

    C)            \[\left( \frac{-{{a}^{2}}m}{c},\frac{{{a}^{2}}}{c} \right)\]            

    D)            \[\left( \frac{-{{a}^{2}}c}{m},\frac{{{a}^{2}}}{m} \right)\]

    Correct Answer: C

    Solution :

               Find points of intersection by simultaneously solving for x and y from \[y=mx+c\] and \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\] which comes out as \[\left( -\frac{{{a}^{2}}m}{c},\ \frac{{{a}^{2}}}{c} \right)\].


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