Question

The circle below is centered at the point (3, 1) and has a radius of length 2.

What is its equation?

Answer 1

Explanation:

an equation of the circle Center at the w(a,b) and ridus : r is :

(x-a)² +(y-b)² = r²

in this exercice : r = 2 and a =3 b =1

so : (x-3)² +(y-1)² = 2²

Answer 2

`$ x^2 + y^2 - 3x -y +6 = 0 $`

Explanation:

The equation of a circle with center
`$ (h,k) $` and radius
`$ r$` is given by

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

Here the center is:
`$ (3,1) $` and radius is
`$ 2 $`.

Therefore, we have:

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

`$ \implies x^2 -3x +9 +y^2 -y +1 = 4 $`

`$ \implies x^2 + y^2 -3x -y +6 = 0 $`