Circumcenter of triangle

The point of intersection of the perpendicular bisectors of the sides of a triangle is called its circumcenter.

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Note: The perpendicular bisectors of the sides of a triangle may not necessarily pass through the vertices of the triangle.

Circumcenter is equidistant to all the three vertices of a triangle.

The circumcenter is the centre of the circumcircle of that triangle.

Circumcenter is denoted by O (x, y).

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The circumcenter of an acute angled triangle lies inside the triangle.

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The circumcenter of the right-angled triangle lies at the midpoint of the hypotenuse of the triangle.

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The circumcenter of the obtuse angled triangle lies outside the triangle.

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The circumcenter of any triangle can be constructed by drawing the perpendicular bisector of any of the two sides of that triangle.

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Following are the Steps to Locate the Circumcenter of the Triangle.

Step:1 Draw the perpendicular bisector of any two sides of the given triangle.

Step:2 Extend the perpendicular bisectors until they intersect each other.

Step:3 Mark the intersecting point as O which will be the circumcenter of the triangle.

Finding the third perpendicular bisector will ensure more accuracy of the location of the circumcenter.

Following are the steps to construct the circumcircle to a given triangle.

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Step:1 Locate the circumcenter by constructing the perpendicular bisectors of at least two sides of the given triangle.

Step:2 Place the compass point on the circumcenter O and stretch to any one of the vertices of the given triangle.

Step:3 Rotate compass to draw a circumcircle.

If A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of the ∆ ABC and A, B, C are their respective angles. Then,

\[O\left( {x,y} \right) = \left( {\frac{{{x_1}\sin 2A + {x_2}\sin 2B + {x_3}\sin 2c}}{{\sin 2A + \sin 2B + \sin 2C}},\frac{{{y_1}\sin 2A + {y_2}\sin 2B + {y_3}\sin 2c}}{{\sin 2A + \sin 2B + \sin 2C}}} \right)\]

METHOD:1

Following are the steps to calculate the circumcenter of a given triangle

Step:1 Find the coordinates of midpoint (xm, ym), of sides AB, BC and AC, using the mid-point theorem.

Step:2 Calculate the slope of each side. If the slope of any side is ‘m1’ then, the slope of the line perpendicular to it will be 1/’m1’. Assume, m = 1/’m1’.

Step:3 Using coordinates of midpoint (xm, ym), and the slope of perpendicular line ‘m’. write the equation of line (y-ym) = m (x-xm).

Step:4 Similarly, find the equation of other lines.

Step:5 Solve them to find out their intersection point.

The obtained intersection point will be the circumcenter of the given triangle.

METHOD:2

Since, we know the property of circumcenter that, Circumcenter is equidistant to all the three vertices of a triangle.

Let O (x, y) be the circumcenter of ∆ ABC. Then, the distances to O from the vertices are all equal, we have AO = BO = CO = Circumradius.

Assume that D1 be the distance between the vertex A (x1, y1) and the circumcenter O (x, y), then

D_{1}^{2} = (x - x1)2 + (y - y1)2 ( Using distance formula of two points in a coordinate)

Similarly,

\[D_2^2 = {(x - {x_2})^2} + {(y - {y_2})^2}\]

\[D_3^2 = {(x - {x_3})^2} + {(y - {y_3})^2}\]

Since, D1 = D2 and D2 = D3,

By this we will get two linear equations:

Equation:1

(x - x1)2 + (y - y1)2 = (x – x2)2 + (y – y2)2

Equation:2

(x – x2)2 + (y – y2)2 = (x – x3)2 + (y – y3)2

By solving these two linear equations using a substitution or elimination method, the coordinates of the circumcenter O (x, y) can be obtained.

Solved Examples:

Q.1. Find the circumcenter of ∆ ABC with vertices A = (1, 4), B = (-2, 3), C = (5, 2).

Ans.

Since the distances to the circumcenter O from the vertices are all equal.

So, AO=BO=CO.

From the first equality, we have AO2 = BO2

(x - 1)2 + (y - 4)2 = (x + 2)2 + (y - 3)2

⇒ -2x + 1 – 8y + 16 = 4x + 4 – 6y + 9

⇒ 3x + y = 2 (1)

Similarly, from the second equality, we have BO2 = CO2

⇒ (x + 2)2 + (y - 3)2 = (x - 5)2 + (y - 2)2

⇒ 4x + 4 – 6y + 9 = -10x + 25 – 4y + 4

⇒ 7x – y = 8 (2)

Solving equations (1) and (2)

Adding (1) + (2) gives:

x = 1 which in turn gives y = −1

Therefore, the circumcenter of triangle ABC is O = (1, -1).

Q.2. Using the circumcenter formula, find the circumcenter of ∆ ABC whose vertices A (0, 2), B (0, 0) and C (2, 0) and respective measures of angles A, B and C are 450, 900 and 450.

Ans.

We know that, if A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of the ∆ ABC and A, B, C are their respective angles. Then,

Circumcenter = \[O\left( {x,y} \right) = \left( {\frac{{{x_1}\sin 2A + {x_2}\sin 2B + {x_3}\sin 2c}}{{\sin 2A + \sin 2B + \sin 2C}},\frac{{{y_1}\sin 2A + {y_2}\sin 2B + {y_3}\sin 2c}}{{\sin 2A + \sin 2B + \sin 2C}}} \right)\]

On putting the corresponding values of coordinates of vertices and angle measures of the ∆ ABC in the above formula. We get:

\[O\left( {x,y} \right) = \left( {\frac{{0\sin 2\left( {45} \right) + 0\sin 2\left( {90} \right) + 2\sin 2\left( {45} \right)}}{{\sin 2\left( {45} \right) + \sin 2\left( {90} \right) + \sin 2\left( {45} \right)}},\frac{{2\sin 2\left( {45} \right) + 0\sin 2\left( {90} \right) + 0\sin 2\left( {45} \right)}}{{\sin 2\left( {45} \right) + \sin 2\left( {90} \right) + \sin 2\left( {45} \right)}}} \right)\]

\[O\left( {x,y} \right) = \left( {\frac{{2\left( {\sin 90} \right)}}{{\sin \left( {90} \right) + \sin \left( {180} \right) + \sin \left( {90} \right)}},\frac{{2\left( {\sin 90} \right)}}{{\sin \left( {90} \right) + \sin \left( {180} \right) + \sin \left( {90} \right)}}} \right)\]On putting the values of corresponding trigonometric ratios:

\[\sin {45^o}\frac{1}{{\sqrt 2 }}\] Sin900 = 1 Sin1800 = 0

⇒ O (x, y) =\[\left( {\frac{2}{{1 + 0 + 1}},\frac{2}{{1 + 0 + 1}}} \right)\]

⇒ O (x, y) = \[\left( {\frac{2}{2},\frac{2}{2}} \right)\]

⇒ O (x, y) = (1, 1)

Q.3. Find the circumcenter of ∆ ABC with the vertices A= (1, 2), B= (3, 6) and C= (5, 4).

Ans.

To calculate the coordinates of circumcenter of the ∆ ABC, we have to solve any two bisector equations and then, find out the intersection points that will give the coordinates of the circumcenter.

So, the midpoint of side AB =\[\left( {\frac{{1 + 3}}{2},\frac{{2 + 6}}{2}} \right) = \left( {2,4} \right)\]

And slope of AB = \[\frac{{6 - 2}}{{3 - 1}} = 2\]

The slope of the perpendicular bisector of side AB is negative reciprocal of the slope of AB

So, slope of the perpendicular bisector of side AB = \[ - \frac{1}{2}\]

The Equation of perpendicular bisector of AB with slope \[ - \frac{1}{2}\]and the coordinates (2,4) is,

\[\left( {y - 4} \right) = - \frac{1}{2}\left( {x - 2} \right)\]

x +2y = 10 (1)

Similarly, we will proceed for side AC

The midpoint of side AC =\[\left( {\frac{{1 + 5}}{2},\frac{{2 + 4}}{2}} \right) = \left( {3,3} \right)\]

And slope of AC = \[\frac{{4 - 2}}{{5 - 1}} = \frac{1}{2}\]

So, slope of the perpendicular bisector of side AC = -2

The Equation of perpendicular bisector of AC with slope -2 and the coordinates (3, 3) is,

(y – 3) = -2 (x - 3)

2x + y = 9 (2)

On solving equations (1) and (2), we get:

x =\[\frac{8}{3}\] and

y = \[\frac{{11}}{3}\]

So, the circumcenter of the ∆ ABC is O\[\left( {\frac{8}{3},\frac{{11}}{3}} \right)\].