The Circumcenter, Circumcircle and Circumradius of a triangle

The circumcenter of a triangle is the point of intersection of the perpendicular bisectors of that triangle.

In the following triangle \(▵ABC\),
\(AK\), \(BL\) and \(CJ\) are the perpendicular bisectors of the triangle.
They intersect at the point \(O\), which is the circumcenter of \(△ABC\).

The circumcenter point of a triangle is equidistant from all the vertices of that triangle.

Circumcenter of a triangle

What are a circumcircle and a circumradius?

The circumcircle of a triangle is a circle that passes through all the three vertices of that triangle.
The center of a circumcircle is known as the circumcenter, while radius of the circumcircle is called the circumradius.

A circumcircle is also known as the circumscribed circle.

What is a perpendicular bisector?

A perpendicular bisector is a line that bisects another line in two equal parts and makes an angle of \(90°\) at the point of intersection. It is a perpendicular on the midpoint of another line.

In the above, \(N\) is the midpoint of \(OP\).
So, \(MN\) has bisected \(OP\) in two equal parts.
Again, \(MN\) is also perpendicular to \(OP\).
So, \(MN\) is the perpendicular bisector to \(OP\).

Properties of a circumcenter

• ✓ Circumcenter is the point of intersection of the perpendicular bisectors.
• ✓ It is the centre of the circumcircle of the triangle.
• ✓ All the vertices of the triangle are equidistant from the circumcenter.
• ✓ For an acute triangle, it lies inside the triangle.
• ✓ For an obtuse triangle, it lies outside the triangle.
• ✓ For a right-angled triangle, it lies on the midpoint of the hypotenuse.
• ✓ In an equilateral triangle, the centroid, the orthocenter,
•  the circumcenter and the incenter coincide on the same point.

Circumcenter of an Acute triangle

\(OJ\), \(OK\) and \(OL\) are the perpendicular bisectors.
\(O\) is the circumcenter of the triangle.
For an acute triangle, the circumcenter lies inside the triangle.
\(OA\), \(OB\) and \(OC\) are the circumradii.
The circle is the circumcircle of the triangle.
Note that the triangle is isosceles too.

Circumcenter of an Obtuse triangle

For an obtuse triangle, the circumcenter lies outside the triangle.

Circumcenter of a Right-angled triangle

For an right-angled triangle, the circumcenter lies on the midpoint of the hypotenuse of the triangle, and the hypotenuse becomes the diameter of the circumcircle.

Coordinates of a circumcenter point

If coordinates of the vertices \(A\), \(B\) and \(C\) are \( (-18, 9) \), \( (11, 9) \) and \( (-9, 33) \) respectively, then the coordinates of the circumcenter \(O(x, y)\) can be calculated with the following steps:

Coordinates of \(O = (x, y) \)
Using distance formula \(\; \sqrt{(y_2-y_1)^2 + (x_2-x_1)^2} \) ,
Distance O to A = \(\; OA = \sqrt{(y - 9)^2 + (x - (-18))^2} \)
Distance O to B = \(\; OB = \sqrt{(y - 9)^2 + (x - 11)^2} \)
Distance O to C = \(\; OC = \sqrt{(y - 33)^2 + (x - (-9))^2} \)

Because \(OA\), \(OB\) and \(OC\) are radii of the same circle, we can solve for \(x\) and \(y\) from the above information.

\begin{align} & OA = OB \quad\color{gray}\text{[Being the radii of same circle]} \\\\ \Rightarrow\; & \sqrt{(y - 9)^2 + (x - (-18))^2} = \sqrt{(y - 9)^2 + (x - 11)^2} \\\\ \Rightarrow\; & (y - 9)^2 + (x + 18)^2 = (y - 9)^2 + (x - 11)^2 \\\\ \Rightarrow\; & (x + 18)^2 = (x - 11)^2 \\\\ \Rightarrow\; & x^2 + 2\cdot x\cdot 18 + 18^2 = x^2 - 2\cdot x\cdot 11 + (11)^2 \\\\ \Rightarrow\; & x^2 + 36x + 324 = x^2 - 22x + 121 \\\\ \Rightarrow\; & 36x + 22x = 121-324 \\\\ \Rightarrow\; & 48x = -203 \\\\ \therefore\; & x = -3.5 \\\\ \end{align}

\begin{align} & OB = OC \quad\color{gray}\text{[Being the radii of same circle]} \\\\ \Rightarrow\; & \scriptsize \sqrt{(y - 9)^2 + (x - 11)^2} = \sqrt{(y - 33)^2 + (x - (-9))^2} \\\\ \Rightarrow\; & \scriptsize (y - 9)^2 + (x - 11)^2 = (y - 33)^2 + (x + 9)^2 \quad\color{gray}\text{[squaring both sides]} \\\\ \Rightarrow\; & \scriptsize y^2 - 2\cdot y\cdot 9 + 9^2 + x^2 - 2\cdot x\cdot 11 + 11^2 = y^2 - 2\cdot y\cdot 33 + 33^2 + x^2 + 2\cdot x\cdot 9 + 9^2 \\\\ \Rightarrow\; & \scriptsize \cancel{y^2} - 18y + \cancel{81} + \cancel{x^2} - 22x + 121 = \cancel{y^2} - 66y + 1089 + \cancel{x^2} + 18x + \cancel{81} \\\\ \Rightarrow\; & 66y - 18y = 18x + 22x - 121+1089 \\\\ \Rightarrow\; & 48y = 40x + 968 \\\\ \Rightarrow\; & 48y = 40\cdot(-3.5) + 968 \quad\color{gray}\text{[putting } x=-3.5\text{]} \\\\ \Rightarrow\; & 48y = -140 + 968 \\\\ \Rightarrow\; & 48y = 828 \\\\ \therefore\; & y = 17.25 \end{align}

\begin{align} & OB = OC \quad\color{gray}\text{[Being the radii of same circle]} \\\\ \Rightarrow\; & \sqrt{(y - 9)^2 + (x - 11)^2} = \sqrt{(y - 33)^2 + (x - (-9))^2} \\\\ \Rightarrow\; & (y - 9)^2 + (x - 11)^2 = (y - 33)^2 + (x + 9)^2 \quad\color{gray}\text{[squaring both sides]} \\\\ \Rightarrow\; & \small y^2 - 2\cdot y\cdot 9 + 9^2 + x^2 - 2\cdot x\cdot 11 + 11^2 = y^2 - 2\cdot y\cdot 33 + 33^2 + x^2 + 2\cdot x\cdot 9 + 9^2 \\\\ \Rightarrow\; & \small \cancel{y^2} - 18y + \cancel{81} + \cancel{x^2} - 22x + 121 = \cancel{y^2} - 66y + 1089 + \cancel{x^2} + 18x + \cancel{81} \\\\ \Rightarrow\; & 66y - 18y = 18x + 22x - 121+1089 \\\\ \Rightarrow\; & 48y = 40x + 968 \\\\ \Rightarrow\; & 48y = 40\cdot(-3.5) + 968 \quad\color{gray}\text{[putting } x=-3.5\text{]} \\\\ \Rightarrow\; & 48y = -140 + 968 \\\\ \Rightarrow\; & 48y = 828 \\\\ \therefore\; & y = 17.25 \end{align}

Solving for \(x\) and \(y\) gives us the coordinates of the circumcenter at \(O(-3.5, 17.25)\)

--- x ---