Coordinate Geometry circle question...

A circle passes through the points (2,0) and (8,0) and has the y axis as a tangent. Find the two possible equations.
Thanks.

The x-coordinate of the center is 5 (the average of 2 and 8).
The y-axis is tangent, so radius = 5.
General equation of the circle:
(x - 5)² + (y - k)² = 25

Plug in (2, 0), then solve for k:
(2 - 5)² + (0 - k)² = 25
k² = 16
k = ±4
The equation becomes
(x - 5)² + (y - 4)² = 25
and
(x - 5)² + (y + 4)² = 25

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Start with the equation of a circle:
(x -h)^2 + (y - k)^2 = r^2

dont forget that h and k are the points for the centre of the circle aswell at (h,k).

remember it passes these points:
(2,0) and (8,0)
and as the circle just touches the y-axis, so h = k. (the centre of the circle is the radius distance away from the y axis)

so we can get 3 equations:

(2 - h)^2 + k^2 = r^2 (1)
(8 - h)^2 + k^2 = r^2 (2)
h = k (3)

Now its just a case of rearranging to solve:

equations (1) and (2) must equal each other, and the k term can subtract out:
(2 - h)^2 = (8 - h)^2

now expand and solve to get:
h = 5;

substitute into equation (3) to get:
r = 5;

ssubstitute both into equation (1) or (2) to get k:
k = 4 or -4,

and that gives the constants for both the equations.

so eq 1) (x - 5)^2 + (y - 4)^2 = 5^2
and
eq 2) (x - 5)^2 + (y + 4)^2 = 5^2

let the equation be (X-x1)^2+(Y-y1)^2=R^2
Since y-axis is a tangent, we have (R,y1) as the centre giving the equation,
(X-R)^2+(Y-y1)^2=R^2
(2,0) and (8,0) satisfy this.
(2-R)^2+y1^2=R^2 and (8-R)^2+y1^2=R^2, giving (2-R)^2=(8-R)^2 or R=5
Substituting this value of R in the earlier equation, we get,
9+y1^2=25
y1^2=16
y1= +4 or -4.
Thus the two equations of the circles are
(X-5)^2+(Y-4)^2=25 and (X-5)^2+(Y+4)^2=25.