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Math Help - Help Need in Help With finding centre of circle

  1. #1
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    Help Need in Help With finding centre of circle

    Hi Guys !

    I am currently stuck with this question :

    A Circle and the curve y^2=kx have a common tangent at the point B. The Centre G of the circle lies on the Line AC. Find

    i) The Coordinates of C
    ii) equation of the circle

    Whereby A = (1,-2) ; B = (4,4) and C = (9,6) and k = 4.

    Need some guidance in this question !

    Realli killing tons of my brain cells thinking about this question man !

    Appreciate your help

    Thanks !
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  2. #2
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    Sorry make a typo mistake.

    Question Should be

    Find Coordinates of G instead.
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  3. #3
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    Quote Originally Posted by liukawa View Post
    Hi Guys !

    I am currently stuck with this question :

    A Circle and the curve y^2=kx have a common tangent at the point B. The Centre G of the circle lies on the Line AC. Find

    i) The Coordinates of C
    ii) equation of the circle

    Whereby A = (1,-2) ; B = (4,4) and C = (9,6) and k = 4.

    Need some guidance in this question !

    Realli killing tons of my brain cells thinking about this question man !

    Appreciate your help

    Thanks !
    Firstly, formulate the fact that the centre of the circle (h, k) is on AC.

    Therefore, write the equation of AC by first calculating the slope,
    then using A or C to write the equation.
    Replace x and y with h and k.
    This relates the centre co-ordinates.

    You need 2 equations to solve for 2 unknowns, hence a second relationship is needed.

    The centre is also on the line going from B to (h, k).
    It's slope is perpendicular to the tangent.
    Hence, first write the slope of the tangent and write the equation of the 2nd centreline in terms of h and k.
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  4. #4
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    So you want y^2= kx and (x- x_0)^2+ (y- y_0)^2= R^2 to have a common tangent at (4, 4). One thing that means is that each curve must go through (4, 4). In particular, 4^2= 16= k(4) so k= 4 as you were apparently given, and the parabola is given by y^2= 4x. Then 2y y'= 4 and at y= 4, that is 8y'= 4 or y'= \frac{1}{2}.
    In order that the circle (x- x_0)^2+ (y- y_0)^2= R^2, which has derivative given by 2(x- x_0)+ 2(y- y_0)y'= 0, have the same tangent at (4, 4) we must have (4- x_0)^2+ (4- y_0)^2= R^2 as well as 2(4- x_0)+ 2(4- y_0)\frac{1}{2}= 0 or 2x_0+ y_0= 12. The line through AC, that is, the line through (1, -2) and (9, 6) has slope (6-(-2))/(9- 1)= 8/8= 1 so the line itself is y= 1(x-1)-2= x- 3. Since the center of the circle lies on that line, we must have y_0= x_0- 3.

    That is, we have three equations, (4- x_0)^2+ (4- y_0)^2= R^2, 2x_0+ y_0= 12, and y_0= x_0- 3. It should be easy to solve the last two linear equations for x_0 and y_0. You could, if you wish, put those values into the first equation and solve for R but apparently the problem only asks for the coordinates of the center of the circle.
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  5. #5
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    Problem is simplified by using the formula for the equation of tangent to a parabola at a given point

    parabola Y^2 =4ax
    equation of tangent=2ax/y1 +2ax1/y1 at x1,y1

    bjh
    Last edited by bjhopper; November 28th 2010 at 11:37 AM. Reason: insert correction in equation
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  6. #6
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    Maybe simplest is to use the slope of the tangent to get the equation of a 2nd centreline.
    (for the first equation see earlier post)

    y^2=4x\Rightarrow\ 2y\frac{dy}{dx}=4\Rightarrow\ \frac{dy}{dx}=\frac{4}{2y}

    At the point B(4,4), the tangent slope is \frac{4}{(2)4}=\frac{1}{2}

    which means that the perpendicular circle centreline passing through (4,4) has slope -2 and contains (h,k).

    Use y-y_1=m\left(x-x_1\right) to write the equation, so k-4=-2(h-4).

    Solving the simultaneous equations (centreline equations) finds the centre co-ordinates.
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