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प्रश्न
The centre of a circle is (2a, a – 7). Find the values of a if the circle passes through the point (11, – 9) and has diameter `10sqrt(2)` units.
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उत्तर
By given condition,
Distance between the centre C(2a, a – 7) and the point P(11, – 9), which lie on the circle = Radius of circle
∴ Radius of circle = `sqrt((11 - 2a)^2 + (-9 - a + 7)^2` ...(i) `[∵ "Distance between two points" (x_1, y_1) "and" (x_2, y_2) = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)]`
Given that, length of diameter = `10sqrt(2)`
∴ Length of radius = `"Length of diameter"/2`
= `(10sqrt(2))/2`
= `5sqrt(2)`
Put this value in equation (i), we get
`5sqrt(2) = sqrt((11 - 2a)^2 + (-2 - a)^2`
Squaring on both sides, we get
50 = (11 – 2a)2 + (2 + a)2
⇒ 50 = 121 + 4a2 – 44a + 4 + a2 + 4a
⇒ 5a2 – 40a + 75 = 0
⇒ a2 – 8a + 15 = 0
⇒ a2 – 5a – 3a + 15 = 0 ...[By fractorisation method]
⇒ a(a – 5) – 3(a – 5) = 0
⇒ (a – 5)(a – 3) = 0
∴ a = 3, 5
Hence, the required values of a are 5 and 3.
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Solution: Suppose A(x1, y1) and B(x2, y2)
x1 = –1, y1 = 1 and x2 = 5, y2 = – 7
Using distance formula,
d(A, B) = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`
∴ d(A, B) = `sqrt(square +[(-7) + square]^2`
∴ d(A, B) = `sqrt(square)`
∴ d(A, B) = `square`
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Based on the above information answer the following questions using the coordinate geometry.
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[OR]
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