Question

Find the coordinates of the circumcentre of the triangle whose vertices are (8, 6), (8, -2) and (2, -2). Also find its circumradius.

Answer 1

Hint: To solve this problem we need to know that the circumcentre of the triangle is equidistant from all the three vertices of the triangle. So we will assume any random point P(x, y) such that it is equidistant from all of the three vertices of the triangle. Now if we suppose given coordinates of the triangle to be A, B and C then PA = PB = PC. Now solving first PA = PB and then PB = PC using the distance formula, we will get the required coordinates of the circumcentre. And then the distance between the circumcentre and any one point of the triangle will be its circumradius.

Complete step by step answer:
We are given the three coordinates of the triangle as (8, 6), (8, -2) and (2, -2),
We know that the circumcentre of the triangle is equidistant from al the three vertices of the triangle so we will assume some point say P(x, y) which is equidistant from all the coordinates of the triangle i.e. A(8, 6), B(8, -2) and C(2, -2),
Now we have,
PA = PB = PC
First we will consider PA = PB,
Applying the distance formula on coordinates P(x, y) and A(8, 6), and P(x, y) and B(8, -2) we get
$\sqrt{{{\left( x-8 \right)}^{2}}+{{\left( y-6 \right)}^{2}}}=\sqrt{{{\left( x-8 \right)}^{2}}+{{\left( y+2 \right)}^{2}}}$
Squaring both sides we get,
${{\left( x-8 \right)}^{2}}+{{\left( y-6 \right)}^{2}}={{\left( x-8 \right)}^{2}}+{{\left( y+2 \right)}^{2}}$
Cancelling out ${{\left( x-8 \right)}^{2}}$ from both the sides we get
${{\left( y-6 \right)}^{2}}={{\left( y+2 \right)}^{2}}$
Expanding the squares on both sides,
$\begin{align}
  & {{y}^{2}}+36-12y={{y}^{2}}+4+4y \\
 & 16y=32 \\
 & y=2\,\,....\left( 1 \right) \\
\end{align}$
Now Applying the distance formula on coordinates P(x, y) an B(8, -2) and P(x, y) and C(2, -2) we get,
$\sqrt{{{\left( x-8 \right)}^{2}}+{{\left( y+2 \right)}^{2}}}=\sqrt{{{\left( x-2 \right)}^{2}}+{{\left( y+2 \right)}^{2}}}$
Squaring both sides we get,
${{\left( x-8 \right)}^{2}}+{{\left( y+2 \right)}^{2}}={{\left( x-2 \right)}^{2}}+{{\left( y+2 \right)}^{2}}$
Cancelling out ${{\left( y+2 \right)}^{2}}$ from both sides we get,
$\begin{align}
  & {{\left( x-8 \right)}^{2}}={{\left( x-2 \right)}^{2}} \\
 & {{x}^{2}}+64-16x={{x}^{2}}+4-4x \\
 & 12x=60 \\
 & x=5\,\,....\left( 2 \right) \\
\end{align}$
From equation 1 and 2 we get the coordinates of the circumcentre of the triangle as (5, 2). So, figure will be like,

Now to find its circumradius we will find the distance of circumcentre from any of the three coordinates of the triangle,
Circumradius = distance between (5, 2) and point A(8, 6)
$=\sqrt{{{\left( 5-8 \right)}^{2}}+{{\left( 2-6 \right)}^{2}}}=\sqrt{9+16}=5$

Hence value of circumradius is 5 units.

Note: You need to remember that distance of circumcentre from any three coordinates of the triangle is equal to each other and also that distance is known as circumradius of the triangle. And also keep in mind that sometimes in order to solve a for a particular value we have to assume that value first to get the simplified equation like in his question we assumed the coordinates of the circumcentre as P(x, y).