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

  • question_answer
    The area of the triangle formed by the tangent at    (3, 4) to the circle \[{{x}^{2}}+{{y}^{2}}=25\]and the co-ordinate axes is  [Pb. CET 2004]

    A)            \[\frac{24}{25}\]                     

    B)            0

    C)            \[\frac{625}{24}\]                   

    D)            \[-\left( \frac{24}{25} \right)\]

    Correct Answer: C

    Solution :

                The equation of the tangent at \[P(3,\,4)\] to the circle\[{{x}^{2}}+{{y}^{2}}=25\] is \[3x+4y=25\], which meets the co-ordinate axes at \[A\left( \frac{25}{3},\,0 \right)\] and \[B\,\left( 0,\,\frac{25}{4} \right)\]. If O be the origin, then the \[\Delta OAB\] is a right angled triangle with \[OA=25/3\] and \[OB=25/4\].            Area of the \[\Delta OAB=\frac{1}{2}\times OA\times OB\]                                          = \[\frac{1}{2}\times \frac{25}{3}\times \frac{25}{4}\]=\[\frac{625}{24}\].


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