RAJASTHAN ­ PET Rajasthan PET Solved Paper-2001

  • question_answer
    Radical axis of the circles \[3{{x}^{2}}+3{{y}^{2}}-7x+8y+11=0\]and \[{{x}^{2}}+{{y}^{2}}-3x-4y+5=0\]is

    A)  \[x+10y=2\]      

    B)  \[x+10y+2=0\]

    C)  \[x+10y=8\]      

    D)  \[x+10y+8=0\]

    Correct Answer: A

    Solution :

     Let \[{{S}_{1}}\equiv 3{{x}^{2}}+3{{y}^{2}}-7x+8y+11=0\] \[\Rightarrow \] \[{{S}_{1}}\equiv {{x}^{2}}+{{y}^{2}}-\frac{7}{3}x+\frac{8}{3}y+\frac{11}{3}=0\] and\[{{S}_{2}}\equiv {{x}^{2}}+{{y}^{2}}-3x-4y+5=0\] Radical axis, \[{{S}_{1}}-{{S}_{2}}=0\] \[\Rightarrow \] \[\left( {{x}^{2}}+{{y}^{2}}-\frac{7}{3}x+\frac{8}{3}y+\frac{11}{3} \right)\] \[-({{x}^{2}}+{{y}^{2}}-3x-4y+5)=0\] \[\Rightarrow \]\[-\frac{7}{3}x+3x+\frac{8}{3}y+4y+\frac{11}{3}-5=0\] \[\Rightarrow \] \[\frac{2}{3}x+\frac{20}{3}y-\frac{4}{3}=0\] \[\Rightarrow \] \[2x+20y-4=0\] \[\Rightarrow \] \[x+10y-2=0\]


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