Plotting the diagram of the given data we get,

Now to find the incentre we must calculate distance of sides so we have $AC=10, BC=8 \& AB=6$

So, incetre will be $\left(\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c},\frac{{\displaystyle a{y}_1+b{y}_2+c{y}_3}}{{\displaystyle a+b+c}}\right)$

$\Rightarrow I=\left(\frac{-32+20+12}{24},\frac{{\displaystyle 24+30-30}}{{\displaystyle 24}}\right)$

$\Rightarrow I=\left(0,1\right)$

Now we know that circumcentre of right angled triangle is given by midpoint of hypotenuse, so circumcentre $C=\left(-1,-1\right)$

Now distance between circumcentre and incentre is given by $\sqrt{1^2+2^2}=\sqrt{5}$