Question

Please help, try and be quick this is due soon

Answer 1

Given the equation of the circle.

`(x-3)^2+y^2=1`

To find: Plot the center. Then plot a point on the circle.

Solution:

The general expression of the circle is

`(x-h)^2+(y-k)^2=a^2`

Here, the center of the circle is at (3,0) and the radius of the circle is

`\begin{gathered} a^2=1 \\ a=1 \end{gathered}`

Hence, the graph is