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  1. #1
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    circle

    determine the coordinates of the centre and the radius of the circle with equation: 2x^2 + 2y^2 - 16x + 4y = 1

    i started by making it 2[x^2 - 8x + (-4)^2] + 2[y^2 + 2y + (1)^2] = 1 + (-8)^2 + (2)^2 and then

    2(x - 4)^2 + 2(y + 1)^2 = \sqrt69 and i said the coordinates are (8, -1) and the radius \sqrt69

    the book says the radius is \frac {1}{2} \sqrt70 and the coordinates are (4, -1)

    can someone tell me how i'm wrong please? thankyou
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  2. #2
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    Quote Originally Posted by mark View Post
    determine the coordinates of the centre and the radius of the circle with equation: 2x^2 + 2y^2 - 16x + 4y = 1

    i started by making it 2[x^2 - 8x + (-4)^2] + 2[y^2 + 2y + (1)^2] = 1 + (-8)^2 + (2)^2 and then

    2(x - 4)^2 + 2(y + 1)^2 = \sqrt69 and i said the coordinates are (8, -1) and the radius \sqrt69

    the book says the radius is \frac {1}{2} \sqrt70 and the coordinates are (4, -1)

    can someone tell me how i'm wrong please? thankyou
    <br />
2x^2 + 2y^2 - 16x + 4y = 1<br />

    2(x^2+y^2-8x+2y) = 1

    (x^2+y^2-8x+2y) = \frac{1}{2}

    (x-4)^2 - 16 + (y+1)^2-1 = \frac{1}{2}

    (x-4)^2 + (y+1)^2 = \frac{33}{2}<br />

    Centre is (4,-1) and radius is \sqrt{\frac{33}{2}}

    Not sure where \frac{1}{2} \,\sqrt{70} comes from
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  3. #3
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    isn't it \sqrt\frac{35}{2}?
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  4. #4
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    Quote Originally Posted by mark View Post
    isn't it \sqrt\frac{35}{2}?
    A simple typo.
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