How do you find the center of a circle through (3,1) and tangent to the x-axis at (0,0)?

Answer 1

Since it is tangent to the x-axis at (0,0), the center must be on the y-axis, lets say at (0,n)

Then, the distance from (0,n) to (0,0) is the same as the distance from (0,n) to (3,1)

so, n = √[(0-3)^2 + (n-1)^2]

n^2 = 9 + n^2 - 2n + 1
2n-1 = 9
n = 5

So the center is at (0,5)

Answer 2

centre (3,1) and radius r = 1
(x - 3)² + (y - 1)² = 1²
(x - 3)² + (y - 1)² = 1