Question

The coordinates of the vertices ofa triangle are (1,2),(2,3),(3,1).Find the co ordinates of the centre of its circumcircle and the circumradius

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that the triangle is ABC having

• Coordinates of A (1, 2)

• Coordinates of B (2, 3)

• Coordinates of C (3, 1)

Let further assume that P (x, y) be the circumcenter of triangle ABC.

So, P is equidistant from A, B and C respectively.

So, PA = PB = PC

Now, We know

Distance Formula

Let A(x₁, y₁) and B(x₂, y₂) be two points in the coordinate plane, then distance between A and B is given by

$\rm\implies \:\boxed{\tt{ AB = \sqrt{ {(x_{1} - x_{2}) }^{2} + {(y_{2} - y_{1})}^{2} }}}$

Consider,

$\rm :\longmapsto\:PA=PB$

$\rm :\longmapsto\:PA^{2} =PB^{2}$

$\rm :\longmapsto\: {(x - 1)}^{2} + {(y - 2)}^{2} = {(x - 2)}^{2} + {(y - 3)}^{2}$

$\rm :\longmapsto\: {x}^{2} + 1 - 2x + {y}^{2} + 4 - 4y = {x}^{2} + 4 -4x+ {y}^{2} + 9 - 6y$

$\rm :\longmapsto\:1 - 2x - 4y = 9 - 4x - 6y$

$\rm :\longmapsto \: - 2x - 4y + 4x + 6y = 9 - 1$

$\rm :\longmapsto \: 2x + 2y = 8$

$\rm :\longmapsto \: 2(x + y) = 8$

$\bf\implies \:x + y = 4 - - - - (1)$

Now, Consider

$\rm :\longmapsto\:PB=PC$

$\rm :\longmapsto\:PB^{2} =PC^{2}$

$\rm :\longmapsto\: {(x - 2)}^{2} + {(y - 3)}^{2} = {(x - 3)}^{2} + {(y - 1)}^{2}$

$\rm :\longmapsto\: {x}^{2} +4 - 4x + {y}^{2} + 9 - 6y = {x}^{2} +9 -6x+ {y}^{2} + 1 - 2y$

$\rm :\longmapsto\:4 - 4x - 6y = 1 - 6x - 2y$

$\rm :\longmapsto\: - 4x - 6y + 6x + 2y = 1 - 4$

$\rm :\longmapsto\: 2x - 4y = - 3$

$\rm :\longmapsto\: 2(4 - y) - 4y = - 3 \: \: \: \{using \: (1) \}$

$\rm :\longmapsto\: 8 - 2y - 4y = - 3$

$\rm :\longmapsto\: - 6y = - 3 - 8$

$\rm :\longmapsto\: - 6y = - 11$

$\bf\implies \:y = \dfrac{11}{6}$

On substituting the value of y in equation (1), we get

$\rm :\longmapsto\:x + \dfrac{11}{6} = 4$

$\rm :\longmapsto\:x = 4 - \dfrac{11}{6}$

$\rm :\longmapsto\:x = \dfrac{24 - 11}{6}$

$\bf\implies \:x = \dfrac{13}{6}$

So,

$\bf\implies \:Coordinates \: of \: P = \bigg(\dfrac{13}{6}, \: \dfrac{11}{6} \bigg)$

Now, Circumradius is given by

$\rm :\longmapsto\:PA$

$\rm \:  =  \: \sqrt{ {\bigg[\dfrac{13}{6} - 1\bigg]}^{2} + {\bigg[\dfrac{11}{6} - 2\bigg]}^{2} }$

$\rm \:  =  \: \sqrt{ {\bigg[\dfrac{13 - 6}{6} \bigg]}^{2} + {\bigg[\dfrac{11 - 12}{6} \bigg]}^{2} }$

$\rm \:  =  \: \sqrt{ {\bigg[\dfrac{7}{6} \bigg]}^{2} + {\bigg[\dfrac{ - 1}{6} \bigg]}^{2} }$

$\rm \:  =  \: \sqrt{ \dfrac{49}{36} + \dfrac{1}{36} }$

$\rm \:  =  \: \sqrt{ \dfrac{49 + 1}{36}}$

$\rm \:  =  \: \sqrt{ \dfrac{50}{36}}$

$\rm \:  =  \: \sqrt{ \dfrac{5 \times 5 \times 2}{6 \times 6}}$

$\rm \:  =  \: \dfrac{5}{6} \sqrt{2} \: units$

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ADDITIONAL INFORMATION

1. Section formula

Let A(x₁, y₁) and B(x₂, y₂) be two points in the coordinate plane and C(x, y) be the point on the line segment which divides AB internally in the ratio m₁ : m₂, then the coordinates of C will be given by

$\sf\implies\boxed{\tt{ (x,y) = \bigg(\dfrac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}, \dfrac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}\bigg)}}$

2. Mid-point formula

Let A(x₁, y₁) and B(x₂, y₂) be two points in the coordinate plane and let C(x, y) be the mid-point of AB, then the coordinates of C will be given by

$\sf\implies\boxed{\tt{ (x,y) = \bigg(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\bigg)}}$

3. Centroid of a triangle

Centroid of a triangle is defined as the point where the medians of the triangle meet and is represented by the symbol G.

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of a triangle. Let G(x, y) be the centroid of the triangle, then the coordinates of G will be

$\sf\implies \: \boxed{\tt{ (x, y) = \bigg(\dfrac{x_{1}+x_{2}+x_{3}}{3}, \dfrac{y_{1}+y_{2}+y_{3}}{3}\bigg)}}$