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The lines \mathrm{2 x-3 y+5=0} and \mathrm{3 x-4 y=7} are diameters of a circle of area 154 sq. units, then the equation of the circle is 

Option: 1

\mathrm{x^{2}+y^{2}+2 x-2 y-62=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

\mathrm{x^{2}+y^{2}-2 x+2 y-62=0}


Answers (1)

best_answer

The centre of the required circle lies at the intersection of \mathrm{2 x-3 y-5= 0 \: and \: 3 x-4 y-7=0}.Thus, the coordinates of the centre are \mathrm{(1,-1)} let \mathrm{r} be the radius of the circle. Then, by hypothesis

\mathrm{\pi r^{2}=154 \Rightarrow r=7}
Hence, the equation of the required circle is
\mathrm{(x-1)^{2}+(y+1)^{2}=7^{2} \: p \: x^{2}+y^{2}-2 x+2 y-47=0}.

Hence (C) is the correct answer.

 

Posted by

Ritika Jonwal

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