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Question
Find the centre of the circle passing through (6, -6), (3, -7) and (3, 3)
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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 centre of a circle is at equal distance from all the points on its circumference.
Here it is given that the circle passes through the points A(6,−6), B(3,−7) and C(3,3).
Let the centre of the circle be represented by the point O(x, y).
So we have AO = BO = CO
`AO = sqrt((6 - x)^2 + (-6-y)^2)`
`BO = sqrt((3 - x)^2 + (-7 - y)^2)`
`CO = sqrt((3 - x)^2 + (3 - y)^2)`
Equating the first pair of these equations we have,
AO = BO
`sqrt((6 - x)^2 + (-6-y)^2) = sqrt((3 -x)^2 + (3 -y)^2)`
Squaring on both sides of the equation we have,
`(6 - x)^2 + (-6-y)^2 = (3 - x)^2 + (3 - y)^2`
`36 + x^2 - 12x + y^2 + 12y = 9 + x^2 - 6x + y^2 - 6y`
6x - 18y = 54
`x - 3y= 9`
Now we have two equations for ‘x’ and ‘y’, which are
3x + y = 7
x - 3y = 9
From the second equation we have y = 3x + 7. Substituting this value of ‘y’ in the first quation we have,
`x - 3(-3x + 7) = 9`
x + 9x - 21 = 9
10x = 30
x = 3
Therefore the value of ‘y’ is,
y = 3x + 7
= -3(3) + 7
y = -2
Hence the co-ordinates of the centre of the circle are (3, -2).
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