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.
Properties of Circumcenter of 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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Locating Circumcenter of Triangle Through Construction
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.
Construction of Circumcircle to a Given Triangle
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.
Circumcenter formula
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,
\[O\left( {x,y} \right) = \left( {\frac{{{x_1}\sin 2A + {x_2}\sin 2B + {x_3}\sin 2c}}{{\sin 2A + \sin 2B + \sin 2C}},\frac{{{y_1}\sin 2A + {y_2}\sin 2B + {y_3}\sin 2c}}{{\sin 2A + \sin 2B + \sin 2C}}} \right)\]
Methods to Calculate Circumcenter of Triangle
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
D12 = (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\[\left( {\frac{8}{3},\frac{{11}}{3}} \right)\].
