In a triangle ABC with vertices A(1,2), B(2, 3) and C(3,1) and \(\angle C=cos^-1\left(\dfrac{4}{5}\right),\angle B=\angle A=cos^-1\left(\dfrac{1}{\sqrt{10}}\right)\). Then the coordinates of circumcentre of the triangle are:

Concept:

The circumcenter of a triangle is the point where the perpendicular bisectors of the sides of the triangle intersect. The circumcenter is also the center of the circle passing through the three points and the distance of the point from the vertices of the triangle is equal.

Formula used:

Distance of two coordinates (x₁, y₁) and (x₂, y₂) is given by distance formula:

\(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

Calculation:

Here, A(1, 2), B(2, 3) and C(3, 1) are the vertices of the triangle ABC.

Let O(x, y) be the coordinates of the circumcenter of the triangle then according to the definition of circumcenter of triangle,

OA = OB = OC 

or OA² = OB² = OC²

Now, we find the distances.

OA² = (x - 1)² + (y - 2)² = x² - 2x + 1 + y² - 4y + 4 = x² + y² - 2x - 4y + 5 

OB2 = (x - 2)2 + (y - 3)2 = x2 - 4x + 4 + y2 - 6y + 9 = x2 + y2 - 4x - 6y + 13

OC2 = (x - 3)2 + (y - 1)2 = x2 - 6x + 9 + y2 - 2y + 1 = x2 + y2 - 6x - 2y + 10

As,

 OA2 = OB2

x2 + y2 - 2x - 4y + 5 = x2 + y2 - 4x - 6y + 13

⇒ -2x + 4x - 4y + 6y = 13 - 5

⇒ 2x + 2y = 8        .........(i)

Now,

 OB2 = OC2

x2 + y2 - 4x - 6y + 13 = x2 + y2 - 6x - 2y + 10

⇒ -4x + 6x - 6y + 2y = 10 - 13

⇒ 2x - 4y = -3        .........(ii)

Subtracting (ii) from (i) we get

2y + 4y = 8 + 3

6y =11

\(y=\frac{11}{6}\)

Put value of y in equation (i) we get,

\(2x+2(\frac{11}{6})=8\)

\(2x+\frac{11}{3}=8\)

\(2x=\frac{13}{3}\)

\(x=\frac{13}{6}\)

Hence, the coordinates of the triangle are \(\left(\dfrac{13}{6},\dfrac{11}{6}\right)\)​