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  •  The lines and <img alt="\mathrm{3 x-4 y=7}" src="https://ent

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


Option: 2

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


Option: 3

\mathrm{x} 2+\mathrm{y} 2-2 \mathrm{x}+2 \mathrm{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 \mathrm{3 x-4 y-7=0}.
Thus, the coordinates of the centre are \mathrm{(1,-1)} let \mathrm{r} be the \mathrm{\pi \mathrm{r} 2=154}

\mathrm{\Rightarrow \mathrm{r}=7}

Hence, the equation of the required circle is

\mathrm{(x-1)^{2}+(y+1)^{2}=72 \quad x^{2}+y^{2}-2 x-47=0}

Posted by

Pankaj

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