Question

Find the equation of the circum circle of the triangle formed by the lines
$5x-3y+4=0,2x+3y-5=0,x+y=0$

Answer 1

Let $AB\Rightarrow 5x-3y+4=0$ ......... $\left(1\right)$

$BC\Rightarrow 2x+3y-5=0$ ......... $\left(2\right)$

$CA\Rightarrow x+y=0$ ........... $\left(3\right)$

$5x-3y=-4$

$-5x+5y=0$

____________

$-8y=-4\Rightarrow y=\frac{1}{2}$ $\therefore A(-\frac{1}{2},\frac{1}{2})$

$\frac{1}{2}+x=0\Rightarrow x=-\frac{1}{2}$

$2x+3y=5$

$2x+2y=0$

____________

$y=5$ $a=-5$ $\therefore C(-5,5)$

$5x-3y=-4$

$2x+3y=5$

___________

$7x=1\Rightarrow x=\frac{1}{7}$ $\therefore (\frac{1}{7},\frac{11}{7})$

$5\left(\frac{1}{7}\right)-3y=-4\Rightarrow y=\frac{11}{7}$

Equation of circles is $x^2+y^2+2gx+2fy+c=0$ .......... $\left(4\right)$

Putting $A(-\frac{1}{2},\frac{1}{2}),B(\frac{1}{7},\frac{11}{7})$ and $c(-5,5)$ conrequently.

$\frac{1}{4}+\frac{1}{4}-g+f+c=0\Rightarrow -g+f+c=-\frac{1}{2}$ ..... $\left(5\right)$

$\frac{1}{49}+\frac{121}{49}+\frac{2}{7}g+\frac{22}{7}f+c=0\Rightarrow 14g+154f+7c=-122$ .......... $\left(6\right)$

$25+25-10g+10f+c=0\Rightarrow -10g+10f+c=-50$ ....... $\left(7\right)$

$10gg+10f+10c=-5$

$-10g+10f+c=-50$

_________________

$9x=45\Rightarrow c=5$

$-g+f=-\frac{1}{2}-5=-\frac{7}{2}\Rightarrow -2g+2f=-7$

$14g+154f+35=-122\Rightarrow 14g+154f=-157$

$-14g+14f=-49$

$14g+154f=-157$

_______________

$168f=-206\Rightarrow f=\frac{103}{84}g=\frac{397}{84}$

$\therefore$ Equation of circle $\Rightarrow 84x^2+84y^2+794x+206y+420=0$

or $42x^2+42y^2+397c+103y+210=0$