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Math Help - Everone loves a cirlce(Worried)

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    Exclamation Everone loves a cirlce(Worried)

    The equation of a cirlce is (x+1)(squared) + y(squared)=13
    i)Write down the centre and radius of this cirlce
    ii)find the equation of the tangent to this cirlce at the point (2,2)

    Thank you
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by eftheking View Post
    The equation of a cirlce is (x+1)(squared) + y(squared)=13
    i)Write down the centre and radius of this cirlce
    ii)find the equation of the tangent to this cirlce at the point (2,2)

    Thank you
    (x - h)^2 + (y - k)^2 = r^2
    is the general equation of a circle where the center is at (h, k) and the radius is r.

    As to the tangent line, the first derivative is the slope of the tangent line, right? so take the derivative (implicitly):
    (x + 1)^2 + y^2 = 13

    2(x + 1) + 2y \frac{dy}{dx} = 0

    \frac{dy}{dx} = -\frac{x + 1}{y}

    So at (2, 2) the slope of the tangent line is
    \frac{dy}{dx} = -\frac{3}{2}

    So find the line with a slope of -3/2 that passes through the point (2, 2).

    -Dan
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    ok.i got the centre of the circle to be -1 for the x co-ordinate, but becase it is y(squared) i did not no what to do for it.is it simply 0?
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    Bar0n janvdl's Avatar
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    Quote Originally Posted by eftheking View Post
    The equation of a cirlce is (x+1)(squared) + y(squared)=13
    i)Write down the centre and radius of this cirlce
    ii)find the equation of the tangent to this cirlce at the point (2,2)

    Thank you
    Centre: (-1 ; 0)
    Radius: \sqrt{13}

    The tangent must make an angle of 90 degrees to the radius. (Sorry i forgot the proper English word for that)

    So let's work out the equation from the radius to that (2;2) point.

    y_{r} = mx + c

    = \frac{2}{3} x + c

    Solve for c and you'll find:

    y_{r} = \frac{2}{3} x + \frac{2}{3}

    ==============

    Now we must find a line that has a gradient of \frac{-3}{2}

    y_{t} = mx + c

    = \frac{-3}{2} x + c

    Solve for c by substituting the points (2 ; 2)
    [Note: You cannot use (-1;0)!! The line does not pass through that point.]

    y_{t} = \frac{-3}{2} x + 5
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    Bar0n janvdl's Avatar
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    Quote Originally Posted by eftheking View Post
    A road 12m wide runs through a tunnel 18m high. The cross section of the tunnel is a major segment of a circle is seen in the attachment.

    I) calculate the radius, BC, of the cirlce
    i)find the angle ABC
    I)Find the cross sectional area of the tunnel
    On MHF we request that you should please post different questions in different threads.

    (Each question should be in it's own thread).
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  6. #6
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    no bother
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