Question

Find the circumcentre and circumradius of a triangle with vertices are $(3,0),(-1,-6),(4,-1)$.

Answer 1

Let the vertices of the triangle be $A(3,0),B(-1,-6)$ and $C(4,-1)$.

$P(x,y)$ be the required circumcentre and r is the circumradius of the $△$ABC,

Then, we must have,

$r^2=P{A}^2=(x-3{)}^2+(y-0{)}^2$ --------(1)

$r^2=P{B}^2=(x+1{)}^2+(y+6{)}^2$ --------(2)

$r^2=P{C}^2=(x-4{)}^2+(y+1{)}^2$ --------(3)

From (1) and (2), we get,

$(x-3{)}^2+y^2=(x+1{)}^2+(y+6{)}^2$

$\Rightarrow x_2+9-6x+y^2=x^2+1+2x+y^2+36+12y$

$\Rightarrow 9-6x=37+2x+12y$

$\Rightarrow 8x+12y+28=0$

$\Rightarrow 2x+3y+7=0$

$\therefore x=\frac{-3y-7}{2}$ ----------(4)

From, (2) and (3), we get

$(x+1{)}^2+(y+6{)}^2=(x-4{)}^2+(y+1{)}^2$

$\Rightarrow x^2+1+2x+y^2+36+12y=x^2+16-8x+y^2+1+2y$

$\Rightarrow 2x+12y+37=2y-8x+17$

$\Rightarrow 10x+10y+20=0$

$\Rightarrow x+y+2=0$

$\Rightarrow \frac{-3y-7}{2}+y+2=0$

$\Rightarrow -3y-7+2y+4=0$

$\Rightarrow y=-3$

Now, $x+y+2=0$

$\Rightarrow x-3+2=0$

$\Rightarrow x=1$

$\therefore$ Required circumcentre=$(1,3)$

Circumradius$=r=\sqrt{(x-3{)}^2+y^2}$

$\Rightarrow r=\sqrt{4+9}$

$\Rightarrow r=\sqrt{13}$