Circumcenter of a Triangle and Its Formula

In this article, we will cover the concept of Orthocentre. This category falls under the broader category of Coordinate Geometry, which is a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. A total of thirteen questions have been asked on JEE MAINS( 2013 to 2023) from this topic including one in 2018, one in 2019, two in 2021, five in 2022, and four in 2023.

Circumcentre: Definition

The Circumcentre (O) of a triangle is the point of intersection of the perpendicular bisectors of the sides of a triangle. The perpendicular bisector of a side of a triangle is the line through the midpoint of a side and perpendicular to it.

A circumference is also defined as the centre of a circle that passes through the vertices of a given triangle.

The circumcentre of a triangle is the centre of the circumscribing circle of the triangle. Also, it is the point where perpendicular bisectors of sides of a triangle are concurrent. The Circumcentre of the acute-angled triangle lies inside the triangle and that of the obtuse-angled triangle lies outside the triangle. The Circumcentre of the right-angled triangle is the midpoint of the hypotenuse.

Circumcentre of triangle formula

Let A, B, and C be the vertices of triangle ABC, then the coordinates circumcentre is:

Coordinates of Circumcentre (O) is

$\left(\frac{\mathrm{x}_1 \sin 2 \mathrm{~A}+\mathrm{x}_2 \sin 2 \mathrm{~B}+\mathrm{x}_3 \sin 2 \mathrm{C}}{\sin 2 \mathrm{~A}+\sin 2 B+\sin 2 C}, \frac{\mathrm{y}_1 \sin 2 \mathrm{~A}+\mathrm{y}_2 \sin 2 \mathrm{~B}+\mathrm{y}_3 \sin 2 \mathrm{C}}{\sin 2 \mathrm{~A}+\sin 2 B+\sin 2 C}\right)$

Note:

For a right-angled triangle, the circumcenter is the midpoint of the hypotenuse.

Properties of circumcircle and circumcentre

1) We know that the perpendicular from the center of the circle to any chord bisects the chord. Thus, the circumcentre of the triangle is the point of concurrency of perpendicular bisectors of the sides of the triangle.

2) The distance of the circumcentre from all the vertices of the triangle is the same which is equal to the circumradius of the triangle.

3) We know that the angle subtended by a chord of the circle at the center is twice the angle subtended by the same chord at any point on the circumference.

4) Distance of circumcentre from sides of the triangle. If O is the center of the circumcircle and E is the side of the triangle.we have OE = R * cos B and OF = R * cos C

5) The Circumcentre of the acute-angled triangle lies inside the triangle.

6) The Circumcentre of the obtuse-angled triangle lies outside the triangle.

7) The Circumcentre of the right-angled triangle is the midpoint of the hypotenuse.

Construction of Circumcenter of Triangle

To construct the circumcenter of any triangle, we need to draw the perpendicular bisectors of any two sides of the triangle. The following are the steps to construct the circumcenter of a triangle:

Step 1: For the given triangle, draw the perpendicular bisectors of any two sides using a compass.

Step 2: With the help of a ruler, extend the perpendicular bisectors till they intersect at a point.

Step 3: Now, mark the point of intersection, which will be the circumcenter of the given triangle.

Solved Examples Based on Circumcentre

Example 1: Let $\mathrm{C}(\alpha, \beta)$ be the circumcenter of the triangle formed by the lines

$\begin{aligned} & 4 x+3 y=69 \\ & 4 y-3 x=17 \text { and } \\ & x+7 y=61\end{aligned}$

Then $(\alpha-\beta)^2+\alpha+\beta$ is equal to [JEE MAINS 2023]

Solution

$
\begin{aligned}
& 4 x+28 y=244 \\
& 4 x+3 y=69 \\
& -\quad-\quad- \\
& \hline 25 y=175 \\
& y=7, x=12 \\
& \mathrm{~A}(12,7) \\
& -3 x+4 y=17 \\
& \frac{3 x+21 y=183}{25 y=200} \\
& y=8, x=5 \\
& \mathrm{~B}(5,8) \\
& \therefore \text { Circumcenter } \\
& \alpha=\frac{17}{2} \beta=\frac{15}{2} \\
& \left(\frac{17}{2}, \frac{15}{2}\right) \\
& (\alpha-\beta)^2+\alpha+\beta \\
& 1+16=17
\end{aligned}
$

Hence, the answer is 17.

Example 2: If the points $P$ and $Q$ are respectively the circumcenter and the orthocentre of a $\triangle \mathrm{ABC}$, the $\overrightarrow{\mathrm{PA}}+\overrightarrow{\mathrm{PB}}+\overrightarrow{\mathrm{PC}}$ is equal to [JEE MAINS 2023]

Solution

$\begin{aligned} & \overrightarrow{\mathrm{PA}}+\overrightarrow{\mathrm{PB}}+\overrightarrow{\mathrm{PC}}=\vec{a}+\vec{b}+\vec{c} \\ & \overrightarrow{\mathrm{PG}}=\frac{\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\vec{c}}{3} \\ & \Rightarrow \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=3 \overrightarrow{\mathrm{PG}}=\overrightarrow{\mathrm{PQ}}\end{aligned}$

Hence, the answer is $\overrightarrow{P Q}$.

Summary

The orthocenter and circumcenter are fundamental points in triangle geometry, each with distinct properties and roles. The orthocenter, where altitudes intersect, reflects the triangle’s perpendicular relationships, while the circumcenter, at the intersection of perpendicular bisectors, defines the centre of the circumcircle. Knowledge of these concepts enriches understanding of triangle symmetry, circle constructions, and their practical applications in fields ranging from architecture to mathematical problem-solving.