How do I solve the inverse function?

Question

2pi r^{2 +8 pi r-A=0}

Answer 1

2pr^2 +8pir - A = 0

A=pir^2

2pir^2 +8pir -pir^2 = 0

pir^2 + 8pir = 0

r^2 + 8r=0

r(r+8) = 0

set each factor = 0

r= 0 or -8 = the radius

Area = A = pir^2 = 0 or 64pi

it's a degenerate circle that collapsed to a point

or a somewhat peculiar circle with negative radius

graphically, it's the origin (0,0) or

a circle centered at the origin with zero radius

x^2 + y^2 = 0

or

x^2 + y^2 = (-8)^2 = 64

which is graphically identical to

x^2 + y^2 =8^2 =64

corresponding polar equtaions are

r=0

and

r^2=64

or if you want an inverse just switch the two variables to get

2piA^2 +8A -r =0

solve for A and r in each equation

A(r) is the inverse of r(A)

at least inverse relations

whether they're both functions depends

unless you really meant

2pir^(2+8r-A) =0

then there is no real solution

as r raised to any real power may approach zero, but never reach it

try an equation calculator on line