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

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    The locus of centre of the circle which cuts the circles\[{{x}^{2}}+{{y}^{2}}+2{{g}_{1}}x+2{{f}_{1}}y+{{c}_{1}}=0\]and\[{{x}^{2}}+{{y}^{2}}+2{{g}_{2}}x+2{{f}_{2}}y+{{c}_{2}}=0\]orthogonally is [Karnataka CET 1991]

    A)            An ellipse                                 

    B)            The radical axis of the given circles

    C)            A conic                                     

    D)            Another circle

    Correct Answer: B

    Solution :

               Let the circle be\[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\]. This cuts the two given circles orthogonally, therefore                    \[2(g{{g}_{1}}+f{{f}_{1}})=c+{{c}_{1}}\]                                                               ?.(i)                    and \[2(g{{g}_{2}}+f{{f}_{2}})=c+{{c}_{2}}\]                                      ?.(ii)                    Subtracting (ii) from (i), we get                    \[2g({{g}_{1}}-{{g}_{2}})+2f({{f}_{1}}-{{f}_{2}})={{c}_{1}}-{{c}_{2}}\]                    So locus of \[(-g,\ -f)\]is                    \[-2x({{g}_{1}}-{{g}_{2}})-2y({{f}_{1}}-{{f}_{2}})={{c}_{1}}-{{c}_{2}}\]                    or \[2x({{g}_{1}}-{{g}_{2}})+2y({{f}_{1}}-{{f}_{2}})+{{c}_{1}}-{{c}_{2}}=0\],                    which is the radical axis of the given circles.


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