equation of straight line and circle

(1)
The mid-point of ((6+2)/2, (-3+3)/2) = (4, 0)

Slope of AB = (-3-3)/(6-2) = -3/2

Slope of the perpendicular bisector
= -1/(slope of AB)
= 2/3

The perpendicular bisector is with slope 2/3 and passes through (4, 0).
The point-slope form of the perpendicular bisector:
y - 0 = (2/3)(x - 4)
3(y - 0) = 2(x - 4)
3y = 2x - 8
2x - 3y - 8 = 0

=====
(2)
Let (a, b) and r be the centre and the radius of the circle.

The circle touches the x-axis at (3,0),
then the line joining (a, b) and (3, 0) is perpendicular to the x-axis.
Hence, a = 3
and the centre of the circle is (3, b)

The y-intercept is 5.
Therefore, the (0, 5) is a point on the circle.

r = Distance between the centre (3, b) and (3, 0)
r = √[(3 - 3)2 + (b - 0)2]
r = b ...... (1)

r = Distance between the centre (3, b) and (0, 5)
r = √[(3 - 0)2 + (b - 5)2
r = √(9 + b2 - 10b + 25)
r = √(b2 - 10b + 34) ...... (2)

(1) = (2):
√(b2 - 10b + 34) = b
b2 - 10b + 34 = b2
-10b + 34 = 0
10b = 34
b = 3.4

(1):
r = 3.4

Equation of the circle:
(x - a)2 + (y - b)2 = r2
(x - 3)2 + (y - 3.4)2 = (3.4)2
x2 - 6x + 9 + y2 - 6.8y + (3.4)2 = (3.4)2
x2 - 6x + 9 + y2 - 6.8y = 0
x2 + y2 - 6x - 6.8y + 9 = 0
5x2 + 5y2 - 30x - 34y + 45 = 0
=