What is the equation of this semicircle?

Question

I know that the formula of a semicircle is
y= (r^2-x^2)^(1/2). However in the graph that I have the center of the semicircle is at point (4,0) and not at the origen. I just need the equation so I can integrate it.

Answer 1

The equation of a semicircle of radius r and center at the origin is
y = √(r² - x²)

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Now always remember this concept:

If you have a function of x, say
y = f(x)

If you want to "move the whole graph" (or Translate the graph) h units about the x-axis and k units about the y-axis, then the function will be
y - k = f(x - h)

To make it again an explicit function,
y = f(x - h) + k

Here, we notice that we subtract h on all x's in the function, and add k to the resulting function.
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Back to the problem...
y = √(r² - x²)

To translate this equation from a center (0,0) to a center (4,0) we notice that we "moved" the whole semicircle 4 units about the x-axis and 0 units about the y-axis. Thus, we subtract 4 from all x's in the function, and add 0 to the resulting function.
y = √[r² - (x - 4)²] + 0

Therefore, the equation of a semicircle with radius r and center at (4,0) is:
y = √[r² - (x - 4)²]

^_^

Answer 2

Equation Of A Semicircle

Answer 3

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RE:
What is the equation of this semicircle?
I know that the formula of a semicircle is
y= (r^2-x^2)^(1/2). However in the graph that I have the center of the semicircle is at point (4,0) and not at the origen. I just need the equation so I can integrate it.

Answer 4

The x-coordinate is 4, so replace x with (x-4) in the equation.

Answer 5

equation of semicircle remains the same as that of circle except that its domain and range changes.

Answer 6

Just replace "x" with "x-4".
y = [r^2 - (x-4)^2] ^ (1/2)