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प्रश्न
Find the centre of the circle passing through (5, -8), (2, -9) and (2, 1).
उत्तर
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(5,−8), B(2,−9) and C(2,1).
Let the centre of the circle be represented by the point O(x, y).
So we have AO = BO = CO
`AO = sqrt((5 - x)^2 + (-8 - y)^2)`
`BO = sqrt((2 - x)^2 + (-9-y)^2)`
`CO = sqrt((2 - x)^2 + (1 - y)^2)`
Equating the first pair of these equations we have,
AO = BO
`sqrt((5 - x)^2 + (-8 - y)^2) = sqrt((2 - x)^2 + (-9 - y)^2)`
Squaring on both sides of the equation we have,
`(5 -x)^2 + (-8 - y)^2 = (2 - x)^2 + (-9 - y)^2`
`25 + x^2 - 10x + 64 + y^2 + 16y = 4 + x^2 - 4x + 81 + y^2 + 18y`
6x + 2y = 4
3x + y = 2
Equating another pair of the equations we have,
AO =BO
`sqrt((5 - x)^2 + (-8 - y)^2) = sqrt((2 - x)^2 + (-9-y)^2)`
Squaring on both sides of the equation we have,
`(5 - x)^2 + (-8 - y)^2 = (2 - x)^2 + (-9 - y)^2`
`25 + x^2 - 10x + 64 + y^2 + 16y = 4 + x^2 - 4x + 81 + y^2 + 18y`
6x + 2y = 4
3x + y = 2
Equating another pair of the equations we have,
AO = CO
`sqrt((5 -x)^2 + (-8 - y)^2) = sqrt((2 -x)^2 + (1 - y)^2)`
`25 + x^2 - 10x + 64 + y^2 + 16y = 4 + x^2 -4x + 1 + y^2 - 2y`
6x - 8y = 84
x - 3y = 14
Now we have two equations for ‘x’ and ‘y’, which are
3x + y = 2
x - 3y = 14
From the second equation we have y = -3x + 2. Substituting this value of ‘y’ in the first equation we have,
x - 3(-3x + 2) = 14
x + 9x - 6 = 14
10x = 20
x = 2
Therefore the value of ‘y’ is,
y = -3x + 2
= -3(2) + 2
y = -4
Hence the co-ordinates of the centre of the circle are (2, -4)
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