JEE Main & Advanced Mathematics Circle and System of Circles Question Bank System of circles

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    The radical centre of the circles \[{{x}^{2}}+{{y}^{2}}-16x+60=0,\,{{x}^{2}}+{{y}^{2}}-12x+27=0,\] \[{{x}^{2}}+{{y}^{2}}-12y+8=0\] is [RPET 2000]

    A)            (13, 33/4)                                 

    B)            (33/4, -13)

    C)            (33/4, 13)                                 

    D)            None of these

    Correct Answer: D

    Solution :

               \[{{S}_{1}}\]\[\equiv \] \[{{x}^{2}}+{{y}^{2}}-16x+60=0\]                                      .....(i)                    \[{{S}_{2}}\]\[\equiv \]\[{{x}^{2}}+{{y}^{2}}-12x+27=0\]                           ..... (ii)                    \[{{S}_{3}}\]\[\equiv \]\[{{x}^{2}}+{{y}^{2}}-12y+8=0\]                                         ..... (iii)                           The radical axis of circle (i) and circle (ii) is                    \[{{S}_{1}}-{{S}_{2}}=0\,\Rightarrow \,-4x+33=0\]                                           .... (iv)                    the radical axis of circle (ii) and circle (iii) is  \[{{S}_{2}}-{{S}_{3}}=0\]\[\Rightarrow \,\,-12+12y+19=0\]                    .....(v)                    Solving (iv) and (v), we get the radical centre\[\left( \frac{33}{4},\,\frac{20}{3} \right)\].


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