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Equation for a circle in terms of constants A and B

  1. Feb 14, 2010 #1
    [1. The problem statement, all variables and given/known data

    If 0<a<b, find the radius R and center (h,k) of the circle that passes through the points (0,a) and (0,b) and is tangent to the x-axis at a point to the right of the origin.

    2. Relevant equations

    ((x-h)^2) + ((y-k)^2)=R^2 (equation of the circle centered around (h,k))

    3. The attempt at a solution

    I already showed that R=k=(a+b)/2, which accounts for both the radius R and the y-component of the center of the circle. However, I'm having some trouble discovering h, the x-component of the circle center, in terms of a and b.

    I also know that k>h. Is there some way I can use the arc length of a quarter of the circle as a hypotenuse for a right triangle using R and h as the height and base, respectively?
  2. jcsd
  3. Feb 15, 2010 #2
    I don't think you can do anything with that.

    I would look at the triangle formed by (0,a) and (0,b) and the center of the circle. You know all the sides of that one.
  4. Feb 15, 2010 #3


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    Since the circle passes through (0, a) and (0, b), both on the y-axis, then that line is a chord. The perpendicular bisector of any chord passes through the center of the circle so we know that the y component of the center is (a+ b)/2. Also, the distance from that midpoint of that chord to either point is (b-a)/2.

    Let "x" be the x component of the center. The distance, squared, from the center to (0, b) is [itex]x^2+ (b- (b+a)/2)^2= x^2+ (b-a)^2/4[/itex]. Since the circle is tangent to the x-axis, the distance from the center to that point of tangency is just the y-component, (b-a)/2. Squaring that, we must have [itex](b+a)^2/4= x^2+ (b-a)^2/4[/math] since each side is a radius, squared. Solve that for x and the rest is easy.

    Interestingly, the x component of the center turns out to be the geometric mean of a and b!
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