Question

What is the equation of the circle with a center at `(6 ,3 )` and a radius of `1`?

Answer 1

`(x-6)^2+(y-3)^2=1`

Explanation:

The general equation of a circle with radius `r` and center at `(h,k)` is `(x-h)^2+(y-k)^2=r^2`. Since the center of this circle is at `(6,3)` and the radius is `1`, we substitute to get:
`(x-(6))^2+(y-(3))^2=(1)^2`

Finish it off for the final result:
`(x-6)^2+(y-3)^2=1`