Ind the equation of a line perpendicular to y - 3x = – 8 that passes through the point (3, 2). (answer in slope-intercept form)

We have that
Find the equation of a line perpendicular to y - 3x = – 8 that passes through the point (3, 2)
y - 3x = – 8-----> y=3x-8------> the slope m=3

we know that
if two lines are perpendicular 
then
m1*m2=-1
m1=3
so
m2=-1/3

with m=-1/3 and point (3, 2) find the equation of the line
y=mx+b
2=(-1/3)*3+b------> 2=-1+b-----> b=3

the equation of the line is
y=(-1/3)x+3

the answer is
y=(-1/3)x+3

see the attached figure

X^2 - 4x +(4/2^2) + y^2 + 8y + (8/2^2) = -4 + (4/2^2) + (8/2^2)(x + 2)^2 + (y + 4)^2 = -4 + 4 + 16your center is (-2, -4) and the radius is the sqrt of -4 + 4 + 16 which is 4