Question

What is the equation of the circle with a center at `(3 ,1 )` and a radius of `1`?

Answer 1

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

Explanation:

The general form for the equation of a circle with a center at `(h,k)` and radius `r` is

`(x-h)^2+(y-r)^2=r^2`

We know that

`(h,k)rarr(3,1)=>h=3,k=1`
`r=1`

So the equation of the circle is

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

or, slightly more simplified (squaring the `1`):

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

The circle graphed:

graph{((x-3)^2+(y-1)^2-1)((x-3)^2+(y-1)^2-.003)=0 [-2.007, 9.093, -1.096, 4.454]}