Homework Help: Equation for a circle in terms of constants A and B

1. Feb 14, 2010

RaoulDuke

[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. Feb 15, 2010

willem2

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.

3. Feb 15, 2010

HallsofIvy

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 $x^2+ (b- (b+a)/2)^2= x^2+ (b-a)^2/4$. 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!