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The equation of a circle whose centre lies on \mathrm{3 x-y-4=0} and \mathrm{x+3 y+2=0} and has an area 154 square units is
 

Option: 1

\mathrm{ x^2+y^2-2 x+2 y-47=0}


Option: 2

\mathrm{ x^2+y^2+2 x-2 y-47=0}


Option: 3

\mathrm{ x^2+y^2-2 x+2 y+47=0}


Option: 4

 none of these


Answers (1)

best_answer

The centre of the circle is the point of intersection of the given lines i.e. the point (1,-1). If r is the radius of the circle, then its area \mathrm{\pi r^2=154}
\therefore \quad  Equation of the circle is \mathrm{(x-1)^2+(y+1)^2=7^2 }

\Rightarrow \quad x^2+y^2-2 x+2 y=47.

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