Half circle sitting on x-axis (like a cradle); what is the width as a function of y?

Answer 1

1) easiest if you situate the point of tangency with x-axis at the origin

2) Say the radius of the circle is r, the circle has center at (0,r), equation is

(x-0)^2 + (y-r)^2 = r^2 or

(y-r)^2 = r^2 - x^2, sqrt both sides

+- (y-r) = sqrt( r^2 - x^2). Since it sits above x-axis,
y-r = sqrt( r^2 - x^2), or
y = r + sqrt( r^2 - x^2)

Done.

Answer 2

Assume the semicircle is tangent to the x-axis at the point P(x1,y1) and is tangent to the y-axis at the point (0, x1).

Then width = 2(x1^2 -(x1-y)^2)^1/2

Answer 3

you need to know what the equation is for this to be answered.