Question

If the points $A(2,-1,1),B(1,-3,-5)$and$C(3,-4,-4)$ form a triangle, find the circum radius for this triangle.

• A. $\sqrt{41}$
• B. $\frac{\sqrt{41}}{2}$
• C. $\sqrt{35}$
• D. $\sqrt{6}$

Answer 1

The correct option is B

$\frac{\sqrt{41}}{2}$

Let's check what kind of triangle is formed using the given points

$AB=\sqrt{1+4+36}=\sqrt{41}$

$BC=\sqrt{4+1+1}=\sqrt{6}$

$CA=\sqrt{1+9+25}=\sqrt{35}$

We can clearly see that

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

$6+35=41$

So,ABC is a right angled triangle , where AB is the hypotenuse

and it is right angled at C

For this right angled triangle, C will be the orthocentre and midpoint of hypotenuse is the circum centre

Circum radius =$\frac{1}{2}|\left(AB\right)|=\frac{1}{2}|\sqrt{41}|=\frac{\sqrt{41}}{2}$

Circum radius of triangle ACB =$\frac{\sqrt{41}}{2}$