JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2012

  • question_answer
        If \[\tan \frac{B-C}{2}=x\cos \frac{A}{2},\]then\[x\]is equal to

    A)  \[\frac{c-a}{c+a}\]                         

    B)  \[\frac{a-b}{a+b}\]

    C)  \[\frac{b-c}{b+c}\]                                         

    D)  None of these

    Correct Answer: C

    Solution :

                    We have the equation of circles \[{{x}^{2}}+{{(y-1)}^{2}}=9\]                    ...(i) and       \[{{(x-1)}^{2}}+{{y}^{2}}=25\]                  ...(ii) For the circle (i), Centre\[{{C}_{1}}(0,1)\]and radius \[{{r}_{1}}=3\] For the circle (ii), Centre\[{{C}_{2}}(1,0)\]and radius\[{{r}_{2}}=5\] Now,     \[{{C}_{1}}{{C}_{2}}=\sqrt{{{1}^{2}}+{{1}^{2}}}=\sqrt{2}\] and           \[{{r}_{2}}-{{r}_{1}}=2\] Clearly,    \[{{C}_{1}}{{C}_{2}}<{{r}_{2}}-{{r}_{1}}\] Therefore, one circle lies entirely inside the other. Hence, there is no common tangent to the given circles.


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