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  •  The equation of the circle having its centre on the line and passi

 The equation of the circle having its centre on the line \mathrm{x+2 y-3=0} and passing through the point of intersection of the circle \mathrm{x^{2}+y^{2}-2 x-4 y+1=0} and \mathrm{x^{2}+y^{2}-4 x-2 y+4=0} is

Option: 1

\mathrm{x^{2}+y^{2}-6 x+7=0}


Option: 2

\mathrm{x}^{2}+\mathrm{y}^{2}-3 \mathrm{x}+4=0


Option: 3

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


Option: 4

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


Answers (1)

best_answer

The equation of a circle passing through the intersection of two given circles is

\left(\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{x}-4 \mathrm{y}+1\right)+\lambda\left(\mathrm{x}^{2}+\mathrm{y}^{2}-4 \mathrm{x}-2 \mathrm{y}+4\right)=0

\Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{x}\left(\frac{1+2 \lambda}{1+\lambda}\right)-2 y \frac{(2+\lambda)}{1+\lambda}+\left(\frac{1+4 \lambda}{1+\lambda}\right)=0 \quad \ldots(i)

co-ordinates of the centre are \left(\frac{1+2 \lambda}{1+\lambda}, \frac{2+\lambda}{1+\lambda}\right).Since the centre lies on \mathrm{x+2 y-3=0}.

\mathrm{\therefore 1+2 \lambda+4+2 \lambda-3-3 \lambda=0}

\mathrm{\Rightarrow \lambda=-2 / 3}

Putting \mathrm{\lambda=-2 / 3} in  (i), we obtain the required circle
\mathrm{\mathrm{x}^{2}+\mathrm{y}^{2}-6 \mathrm{x}+7=0}.
 

Posted by

Sanket Gandhi

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