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

Sep 17, 2016

Centre at $\left(5 , 0\right)$ Radius = $\frac{1}{4}$

Explanation:

The equation of a circle with centre at point $\left(a , b\right)$ and radius $r$ is:

${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$

Here we are given the equation: ${\left(x - 5\right)}^{2} + {y}^{2} = \frac{1}{16}$
Hence, in this example: $a = 5$, $b = 0$ and ${r}^{2} = \frac{1}{16}$

Therefore the centre of the circle is at $\left(5 , 0\right)$

The radius is $\frac{1}{\sqrt{16}} = \frac{1}{4}$ (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]}