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Two circles \mathrm{x^{2}+y^{2}=6} and \mathrm{x^{2}+y^{2}-6x+8=0} are given .Then the equation of the circle through their points of intersection and the point (1,1) is 

Option: 1

\mathrm{x^2+y^2-6 x+4=0 }


Option: 2

\mathrm{ x^2+y^2-3 x+1=0 }


Option: 3

\mathrm{x^2+y^2-4 x+2=0 }


Option: 4

None of these


Answers (1)

best_answer

Equation of the common chord is \mathrm{S-S'=0} 

\mathrm{\begin{aligned} & \Rightarrow x^2+y^2-6 x+8-x^2-y^2+6=0 \\ & \Rightarrow 3 x-7=0 \end{aligned}}

Equation of any circle passing through the point of
intersection is \mathrm{S+\lambda L=0}

\mathrm{\Rightarrow x^2+y^2-6+\lambda(3 x-7)=0} --------(i)

(i) passing through (1, 1)

\mathrm{\begin{aligned} & \Rightarrow \quad 1+1-6+\lambda(-4)=0 \\ & \therefore \lambda=-1 \end{aligned}} 

Equation of the required circle is \mathrm{x^2+y^2-3 x+1=0}

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