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प्रश्न
The centre of a circle is (−4, 2). If one end of the diameter of the circle is (−3, 7) then find the other end
उत्तर
Let the other end of the diameter B be (a, b)
Mid−point of a line = `((x_1 + x_2)/2, (y_1 + y_2)/2)(-3, 7)`
(−4, 2) = `((-3 + "a")/2, (7 + "b")/2)`
∴ `(-3 + "a")/2` = − 4
−3 + a = − 8
a = – 8 + 3
a = −5
`(7 + "b")/2`
7 + b = 4
b = 4 – 7
⇒ b = – 3
The other end of the diameter is (– 5, – 3).
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Point P is the centre of the circle and AB is a diameter. Find the coordinates of points B if coordinates of point A and P are (2, – 3) and (– 2, 0) respectively.
Given: A`square` and P`square`. Let B (x, y)
The centre of the circle is the midpoint of the diameter.
∴ Mid point formula,
`square = (square + x)/square`
⇒ `square = square` + x
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⇒ y = 3
Hence coordinates of B is (– 6, 3).
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