# 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!