(Edited)

The perpendicular bisector of chord AB, which happens to be the y-axis itself, passes through the centre of the circle. (One look and the line of symmetry is obvious)

So, the x-coordinate of the centre of the circle is 0.

Now, let the centre of the circle be G (0, k).

The radii of the circle has the same value in all directions.

So, radius GA = radius GT
sqrt [(-4 - 0)^2 + (0 - k)^2] = 1 - k
(-4 - 0)^2 + (0 - k)^2 = (1 - k)^2
16 + k^2 = 1 - 2k + k^2
16 = 1 - 2k
15 = -2k
-7.5 = k

and radius of circle = 1 - (-7.5) = 8.5.

Equation of circle is
(x - 0)^2 + [y - (-7.5)]^2 = 8.5^2
x^2 + y^2 + 15y + 56.25 = 72.25
x^2 + y^2 + 15y - 16 = 0

Should show how it is a perpendicular bisector first i.e

①
OA = 0 - (-4) = 4
OB = 4 - 0 = 4

Since OA = OB , the y-axis bisects the chord AB at O.

②Since the y-axis is perpendicular to the x-axis, and AB is part of the x-axis, the y-axis is a perpendicular bisector of AB.

③ State the theorem
④ centre lies on y-axis so x-coordinate is 0 since x = 0 on any point of it

Also, if your k is positive (= 7.5), then your centre should be (0,-k) since the centre is below the x-axis

Otherwise, if your centre is (0,k), then for the first line of the equation, it should be 1 - k instead of 1 - (-k)