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Question
Find the coordinates of the circumcentre of the triangle whose vertices are (3, 0), (-1, -6) and (4, -1). Also, find its circumradius.
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Solution
The distance d between two points `(x_1,y_1)` and `(x_2, y_2)`is given by the formula
`d = sqrt((x_1 - x_2)^2 + (y_1 - y_2)^2)`
The circumcentre of a triangle is the point which is equidistant from each of the three vertices of the triangle.
Here the three vertices of the triangle are given to be A(3,0), B(−1,−6) and C(4,−1)
Let the circumcentre of the triangle be represented by the point R(x, y).
So we have AR = BR = CR
`AR = sqrt((3 - x)^2 + (-y)^2)`
`BR = sqrt((-1-x)^2 + (-6 -y)^2)`
`CR = sqrt((4 -x)^2 + (-1-y)^2)`
Equating the first pair of these equations we have,
AR= BR
`sqrt((3 - x)^2 + (-y)^2) = sqrt((-1-x)^2 +(-6-y)^2)`
Squaring on both sides of the equation we have,
`sqrt((3 - x)^2 + (-y)^2) = sqrt((-1-x)^2 + (-6-y))`
`9 + x^2 - 6x + y^2 = 1 + x^2 + 2x + 36 + y^2 + 12y`
8x + 12y = -28
2x + 3y = -7
Equating another pair of the equations we have,
AR = CR
`sqrt((3 - x)^2 + (-y)^2) = sqrt((4 - x)^2 + (-1 - y)^2)`
Squaring on both sides of the equation we have,
`(3 - x)^2 + (-y)^2 = (4 - x)^2 + (-1 - y)^2`
`9 + x^2 - 6x + y^2 = 16 + x^2 - 8x + 1 + y^2 + 2y`
2x - 2y = 8
x - y = 4
Now we have two equations for ‘x’ and ‘y’, which are
2x + 3y = -7
x - y = 4
From the second equation we have y = x - 4. Substituting this value of ‘y’ in the first equation we have,
2x + 3(x - 4) = -7
2x + 3x - 12 = -7
5x = 5
x= 1
Therefore the value of ‘y’ is,
y = x - 4
= 1 - 4
y = -3
Hence the co-ordinates of the circumcentre of the triangle with the given vertices are (1, -3).
The length of the circumradius can be found out substituting the values of ‘x’ and ‘y’ in ‘AR’
`AR = sqrt((3 - x)^2 + (-y)^2)`
`= sqrt((3 -1)^2 + (3)^2)`
`= sqrt((2)^2 +(3)^2)`
`= sqrt(4 + 9)`
`AR = sqrt13`
Thus the circumradius of the given triangle is `sqrt13` units
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