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BCECE Engineering BCECE Engineering Solved Paper-2011

  • question_answer
    The circle \[{{S}_{1}}\]with centre \[{{C}_{1}}({{a}_{1}},{{b}_{1}})\]and radius \[{{r}_{1}}\]touches externally the circle \[{{S}_{2}}\]with centre \[{{C}_{2}}({{a}_{2}},{{b}_{2}})\]and radius \[{{r}_{2}}.\] If the tangent at their common point passes through the origin, then

    A)  \[(a_{1}^{2}+a_{2}^{2})+(b_{1}^{2}+b_{2}^{2})=r_{1}^{2}+r_{2}^{2}\]

    B)         \[(a_{1}^{2}-a_{2}^{2})+(b_{1}^{2}-b_{2}^{2})=r_{1}^{2}-r_{2}^{2}\]

    C)         \[{{(a_{1}^{2}-{{b}_{2}})}^{2}}+(a_{2}^{2}+b_{2}^{2})=r_{1}^{2}+r_{2}^{2}\]

    D)         \[(a_{1}^{2}-b_{1}^{2})+(a_{1}^{2}+b_{2}^{2})=r_{1}^{2}+r_{2}^{2}\]

    Correct Answer: B

    Solution :

    Given two circles are \[{{S}_{1}}\equiv {{(x-{{a}_{1}})}^{2}}+(y-b_{1}^{2})=r_{1}^{2}\]                 ?(i) and \[{{S}_{2}}\equiv {{(x-{{a}_{2}})}^{2}}+(y-b_{2}^{2})=r_{2}^{2}\]          ?(ii) The equation of the common tangent of these two circles is given by \[{{S}_{1}}-{{S}_{2}}=0\]i.e., \[2x({{a}_{1}}-{{a}_{2}})+2y({{b}_{1}}-{{b}_{2}})+(a_{2}^{2}+b_{2}^{2})\] \[-(a_{1}^{2}+b_{1}^{2})+r_{1}^{2}-r_{2}^{2}=0\] If this passes through the origin, then                 \[(a_{2}^{2}+b_{2}^{2})-(a_{1}^{2}+b_{1}^{2})+r_{1}^{2}-r_{2}^{2}=0\]                 \[\Rightarrow \]\[(a_{2}^{2}-a_{1}^{2})+(b_{2}^{2}-b_{1}^{2})=r_{2}^{2}-r_{1}^{2}\]                 \[\Rightarrow \]\[(a_{1}^{2}-a_{2}^{2})+(b_{1}^{2}-b_{2}^{2})=(r_{1}^{2}-r_{2}^{2})\]


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