Circumcentre of a triangle – Distance Formula

Circumcentre of a triangle

When a circle is passing through all the three vertices of the triangle, it is the circle circumscribing the triangle. This is also termed the circumcircle of that triangle. The Circumcentre is the centre of this circumcircle. 

In the figure shown above, “O” is the circumcentre of △PQR.

Properties of circumcentre of the triangle

1. The distance of all the vertices of the triangle from the circumcentre is equal.
2. On this point, the perpendicular bisectors of the three sides of the triangle meet each other.
3. The circumcentre can lie inside or outside or on the triangle.

• In an acute triangle, the circumcentre always lies inside the triangle.
• In an obtuse triangle, the circumcentre always lies outside the triangle.
• Right angle triangle – Circumcentre lies on the midpoint of the hypotenuse of the triangle.

Finding circumcentre using the distance formula

The formula for the distance between any two points \mathrm{A}\left(x_{a}, y_{a}\right) \text { and } \mathrm{B}\left(x_{b}, y_{b}\right) is given below:

\mathrm{AB}=\sqrt{\left(x_{a}-x_{b}\right)^{2}+\left(y_{a}-y_{b}\right)^{2}}

In △PQR, let the coordinates of the vertices P, Q and R be \left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right) \text { and }\left(x_{3}, y_{3}\right) respectively and the coordinate of the circumcentre O is (a, b).

The distance of all the vertices of the triangle from the circumcentre is equal. We have to follow the following steps to find the coordinates of the circumcentre.

1. First, we have to calculate the distances of all the vertices from the circumcentre 

The distance between \mathrm{P} from \mathrm{O}=\mathrm{PO}=\sqrt{\left(x_{1}-a\right)^{2}+\left(y_{1}-b\right)^{2}}

The distance between Q from O=Q O=\sqrt{\left(x_{2}-a\right)^{2}+\left(y_{2}-b\right)^{2}}

The distance between \mathrm{R} from \mathrm{O}=\mathrm{RO}=\sqrt{\left(x_{1}-a\right)^{2}+\left(y_{1}-b\right)^{2}}

2. Equating PO = QO 

\sqrt{\left(x_{1}-a\right)^{2}+\left(y_{1}-b\right)^{2}}=\sqrt{\left(x_{2}-a\right)^{2}+\left(y_{2}-b\right)^{2}}

\Rightarrow\left(x_{1}-a\right)^{2}+\left(y_{1}-b\right)^{2}=\left(x_{2}-a\right)^{2}+\left(y_{2}-b\right)^{2}

Equation PO = RO

\sqrt{\left(x_{1}-a\right)^{2}+\left(y_{1}-b\right)^{2}}=\sqrt{\left(x_{3}-a\right)^{2}+\left(y_{3}-b\right)^{2}}

\Rightarrow\left(x_{1}-a\right)^{2}+\left(y_{1}-b\right)^{2}=\left(x_{3}-a\right)^{2}+\left(y_{3}-b\right)^{2}

3. Now we have 2 equations as shown above and 2 unknown variables “a” and “b”

We have to solve the above two equations to find the coordinate of the circumcentre O (a, b) of △PQR

Solved Examples

1. Find the circumcentre of △PQR where the  coordinates of the Vertices P, Q and R are (0,0), (0,9) and (9,0) respectively

Given:

- x_{1}=0

- \mathrm{y}_{1}=0

- x_{2}=0

- \quad y_{2}=9

- x_{3}=9

- y_{3}=0

Let the Coordinates of circumcentre be (a,b)

Equating the given values in the equations 

\left(x_{1}-a\right)^{2}+\left(y_{1}-b\right)^{2}=\left(x_{2}-a\right)^{2}+\left(y_{2}-b\right)^{2}

\Rightarrow(0-a)^{2}+(0-b)^{2}=(0-a)^{2}+(9-b)^{2}

\Rightarrow \mathrm{a}^{2}+\mathrm{b}^{2}=\mathrm{a}^{2}+81+\mathrm{b}^{2}-18 \mathrm{~b}

\Rightarrow 81-18 \mathrm{~b}=0

\Rightarrow 18 \mathrm{~b}=81

=\Rightarrow \mathrm{b}=4.5

\left(x_{1}-a\right)^{2}+\left(y_{1}-b\right)^{2}=\left(x_{3}-a\right)^{2}+\left(y_{3}-b\right)^{2}

 \Rightarrow(0-a)^{2}+(0-b)^{2}=(9-a)^{2}+(0-b)^{2}

\Rightarrow \mathrm{a}^{2}+\mathrm{b}^{2}=\mathrm{a}^{2}+81-18 \mathrm{a}+\mathrm{b}^{2}

\Rightarrow 81-18 \mathrm{a}=0

\Rightarrow 18 \mathrm{~b}=81

\Rightarrow \mathrm{a}=4.5

Hence the circumcentre of △PQR is (4.5, 4.5)

2. The circumcentre of △PQR is (0,0). If the coordinates of points P and Q are (1,0) and (0,1), find the possible coordinates of point R.

Given:

- x_{1}=1

- \quad \mathrm{y}_{1}=0

- x_{2}=0

- \quad \mathrm{y}_{2}=1

- \quad \mathrm{a}=0

- b=0

Let the coordinate of R be \left(x_{3}, y_{3}\right)

Equating the given values in the equations 

\left(x_{1}-a\right)^{2}+\left(y_{1}-b\right)^{2}=\left(x_{3}-a\right)^{2}+\left(y_{3}-b\right)^{2}

\Rightarrow(1-0)^{2}+(0-0)^{2}=\left(x_{3}-0\right)^{2}+\left(y_{3}-0\right)^{2}

\Rightarrow x_{3}^{2}+y_{3}^{2}=1

\left(x_{2}-a\right)^{2}+\left(y_{2}-b\right)^{2}=\left(x_{3}-a\right)^{2}+\left(y_{3}-b\right)^{2}

\Rightarrow(0-0)^{2}+(1-0)^{2}=\left(x_{3}-0\right)^{2}+\left(y_{3}-0\right)^{2}

 \Rightarrow x_{3}^{2}+y_{3}^{2}=1

This shows that the values of x3 and y3 are not constant and can be any value satisfying the equation x32+y32=1 except the values (1,0) and (0,1) as these are already the other two points of the triangle