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

  • question_answer
    If line \[ax+by=0\] touches \[{{x}^{2}}+{{y}^{2}}+2x+4y=0\] and is a normal to the circle \[{{x}^{2}}+{{y}^{2}}-4x+2y-3=0\], then value of (a,b) will be [AMU 2005]

    A)            (2, 1)                                         

    B)            (1, -2)

    C)            (1, 2)                                         

    D)            (-1, 2)

    Correct Answer: C

    Solution :

     As the line \[ax+by=0\] touches the circle\[{{x}^{2}}+{{y}^{2}}+2x+4y=0\], distance of the centre (?1, ?2) from the line = radius                    Þ \[\left| \frac{-a-2b}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right|=\sqrt{{{(-1)}^{2}}+{{(-2)}^{2}}}\] Þ \[{{(a+2b)}^{2}}=5({{a}^{2}}+{{b}^{2}})\]                    Þ \[4{{a}^{2}}-4ab+{{b}^{2}}=0\] Þ \[{{(2a-b)}^{2}}=0\]                    \ \[b=2a\].                    Next, \[ax+by=0\] is a normal to\[{{x}^{2}}+{{y}^{2}}-4x+2y-3=0\], the centre (2, ?1) should lie on \[ax+by=0\]                    \ \[2a-b=0\Rightarrow b=2a\]. Hence\[a=1\], \[b=2\].


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