Question

How do you find the center and radius of `(X-5)^2 + Y^2 = 1/16`?

Answer 1

Centre at `(5,0)` Radius = `1/4`

Explanation:

The equation of a circle with centre at point `(a,b)` and radius `r` is:

`(x-a)^2+(y-b)^2 = r^2`

Here we are given the equation: `(x-5)^2+y^2 = 1/16`
Hence, in this example: `a=5`, `b=0` and `r^2=1/16`

Therefore the centre of the circle is at `(5,0)`

The radius is `1/sqrt16 = 1/4` (Since the radius must be positive)

This can be seen on the graph of the positive portion of the circle below.

graph{(1/16 - (x-5)^2)^(1/2) [4.6156, 5.355, -0.057, 0.3128]}