Question

Find the equation of the circle circumscribing the triangle formed by the lines $x+y=6,2x+y=4$ and $x+2y=5$

Answer 1

$x+y=\mathrm{6...}\left(1\right)$

$2x+y=\mathrm{4....}\left(2\right)$

$x+2y=\mathrm{5....}\left(3\right)$

solve (1) and (2)

from (1), $y=6-x$

sub in (2)

$2x+6-x=4$

$x=4-6=-2$

$y=6-x=6-(-2)=8$

$(x,y)=(-2,8)$

Solve (2) and (3)

from (2) $y=4-2x$

sub in (3)

$x+2(4-2x)=5$

$x+8-4x=5$

$-3x-5-8$

$-3x=-3$

$x=1$

$2\left(1\right)+y=4$

$y=+2$

$(1,+2)$

Solve (1) and (3)

$x=5-2y$

sub in (1)

$5-2y+y=6$

$y=-1$

$x-1=6$

$x=6+1$

$x=7$

$(7,-1)$

Equation of circumscribing

${(x-a)}^2+{(y-b)}^2=r^2....\left(4\right)$

Substitute value of $x$ and $y$ in (4)

$(-2,8)$ in (4) $\Rightarrow$ $a^2+b^2+4a-16+68=r^2....\left(5\right)$

$(1,+2)$ in (4) $\Rightarrow$ $a^2+b^2-2a-4b+5=r^2....\left(6\right)$

$(7,-1)$ in (4) $\Rightarrow$ $a^2+b^2-14a+2b+50=r^2.......\left(7\right)$

solve (5) and (7) we get $a-b=-\mathrm{1....}\left(8\right)$

solve (7) and (6) we get $2b-4a=-\mathrm{13...}\left(9\right)$

solve (8) and (9)

we get $a=\frac{15}{2},b=\frac{17}{2}$

Substitute $a$ and $b$ in (5) we get $r^2=90.5$

equation is

${[x-\frac{15}{2}]}^2+{[y-\frac{17}{2}]}^2=90.5$