Question

Show that $(-1,4,-3)$ is the circum-centre of the triangle formed by the points $(3,2,-5),(-3,8,-5)$ and $(-3,2,1)$.

Answer 1

R.E.F image

$(x-3{)}^2+(y-2{)}^2+(z+5{)}^2=(x+3{)}^2+(y+8{)}^2+(z+5{)}^2=(x+3{)}^2+(y-2{)}^2+(z-1{)}^2$

$\therefore x-y+5=0$.......(i)

$y-z-7=0$........(ii)

$x-z-2=0$.........(iii)

(ii)+(iii)

$x+y=2z-9=0$

clearly $(-1,4,-3)$ satisfies straight equations

Hence, $(-1,4,-3)$ is circumference of triangle $ABC$