Question

Find the circumcentre and circumradius of the triangle whose vertices are $(1,1),(2,-1)$ and $(3,2)$

Answer 1

Let the coordinates of the vertices of the triangle are $A(1,1),B(2,-1)$ and $C(3,2)$

Now, Lets find the distance between each of the three points.

We will use the distance formula to find the distance between $AB,BC$ and $AC$

Thus $AB=\sqrt{(2-1{)}^2+(-1-1{)}^2}=\sqrt{5}$

$BC=\sqrt{(3-2{)}^2+(2+1{)}^2}=\sqrt{10}$

$AC=\sqrt{(3-1{)}^2+(2-1{)}^2}=\sqrt{5}$

Now$A{B}^2+A{C}^2=10=B{C}^2$

Thus it form a right angled triangle, right angled at $A.$

Now the circumcentre will be the mid-point of the hypotenuse $BC.$

Thus mid-point of $BC$ is $(\frac{2+3}{2},\frac{-1+2}{2})=(\frac{5}{2},\frac{1}{2}).$

And the circumradius is the half the length of $BC$

$\Rightarrow$ Circumradius $=\frac{\sqrt{10}}{2}$