Question

Find the Circumcentre and Circumradius of the triangle whose verices are $(1,1),(2,-1)$ and $(3,2)$

Answer 1

Step 1. Calculate the distance.

Let the coordinates of the vertices of the triangle are

$A=(1,1),B=(2,-1)$and $C=(3,2)$

$$\begin{gathered} AB=\sqrt{{\left(2-1\right)}^2+{\left(-1-1\right)}^2} \\ =\sqrt{{\left(1\right)}^2+{\left(-2\right)}^2} \\ =\sqrt{1+4} \\ =\sqrt{5} \\ BC=\sqrt{{\left(3-2\right)}^2+{\left(2+1\right)}^2} \\ =\sqrt{{\left(1\right)}^2+{\left(3\right)}^2} \\ =\sqrt{10} \\ AC=\sqrt{{\left(3-1\right)}^2+{\left(2-1\right)}^2} \\ =\sqrt{{\left(2\right)}^2+{\left(1\right)}^2} \\ =\sqrt{5} \end{gathered}$$

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

Therefore it forms a right-angled triangle, right-angled at $A$.

Step 2. Find the Circumcentre.

Since the given vertices form a right angle triangle the circumcentre will be the mid-point of the hypotenuse $BC$.

The mid point of $BC$ is

$\left(\frac{2+3}{2},\frac{-1+2}{2}\right)=\left(\frac{5}{2},\frac{1}{2}\right)$

Therefore, the Circumcentre is $\left(\frac{5}{2},\frac{1}{2}\right)$.

Step 3. Finding the Circumradius.

he circumradius is the half the length of $BC$

So Circumradius is $\frac{\sqrt{10}}{2}$

Hence, the Circumradius is $\frac{\sqrt{10}}{2}$ and the Circumcentre is $\left(\frac{5}{2},\frac{1}{2}\right)$,.