Find the coordinates of the point on the circle x²+y²-12x-4y+30=0 which is nearest the origin.?

Ans: (3,1)

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  • 2 years ago
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    x²+y²-12x-4y+30=0 is a circle centred at (6,2) with a radius of √(6²+2²-30) = √10

    and (6,2) is √(6²+2²) = 2√10 from the origin {twice the length of the radius!}

    So the point on the circle x²+y²-12x-4y+30=0 which is nearest the origin is the midpoint of (0,0) and (6,2) which is (3,1).

  • Find the center of the circle

    x^2 - 12x + y^2 - 4y = -30

    x^2 - 12x + 36 + y^2 - 4y + 4 = -30 + 36 + 4

    (x - 6)^2 + (y - 2)^2 = 10

    The center is at (6 , 2)

    Describe a line that passes through (0 , 0) and (6 , 2)

    m = (2 - 0) / (6 - 0) = 2/6 = 1/3

    y = (1/3) * x

    Find when this intersects the circle

    x^2 + (x/3)^2 - 12x - 4 * (x/3) + 30 = 0

    x^2 + (1/9) * x^2 - 12x - 4x/3 + 30 = 0

    (10/9) * x^2 - (12 + 4/3) * x + 30 = 0

    10 * x^2 - 9 * (12 + 4/3) * x + 9 * 30 = 0

    10x^2 - (108 + 12) * x + 270 = 0

    10x^2 - 120x + 270 = 0

    x^2 - 12x + 27 = 0

    x^2 - 12x = -27

    x^2 - 12x + 36 = -27 + 36

    (x - 6)^2 = 9

    x - 6 = -3 , 3

    x = 3 , 9

    x = 3

    y = x/3

    y = 3/3

    y = 1

    (3 , 1)

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