Ax+y=b and ax^2+y^2=c

solve for y b and a are constants x and y are variables
pls help I did alot of steps and got down to the quadratic formula

ax + y = b (1)
ax² + y² = c (2)

from 1
y = b - ax

substitute into (2)

ax² + (b - ax)² = c
ax² + b² - 2abx + (ax)² = c
(a + a²)x² - 2abx + b² - c = 0
discriminant D = (2ab)² - 4(a + a²)(b² - c)
D = 4[a²b² - (ab² -ac + a²b² - a²c)]
D = 4(a²c + ac - ab²)
√D = 2√(a²c + ac - ab²)

x = [2ab ±2√(a²c + ac - ab²)]/(2(a + a²))
2 canceled out
x = [ab ±√(ac + c - b²)]/(a + a²)

so
y = b - a[ab ±√(ac + c - b²)]/(a + a²)

y = b - [ab ±√(ac + c - b²)]/(a + 1)

If you have b and c, plug in to find x and y
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Wow, all the effort just useless
so you meant the system is
ax² + y² = b²
ax² + y² = c ???

if so subtract one from another

b² = c
b = √c

x and y can be anything.
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edit 2:
ax + y = b
ax² + y² = b²

if this is the problem then

ax² + (b - ax)² = b²
ax² + b² - 2abx + a²x² - b² = 0
(a + a²)x² - 2abx = 0

x[(a + a²)x - 2ab] = 0
x = 0 is one of the root and then y = b

x = 2ab/(a + a²)

x = 2b/(1 + a) then

y = b - 2ab/(1 + a)

ax + y = b
y = (b/ax) divide both sides by ax to get y by itself

ax = b - y subtract y from both sides
x = [(b - y) / a] divide both sides by a

x^2 = [(c - y^2) / a]
x = sqrt[(c - y^2) / a]

y^2 = c - ax^2
y = sqrt(c - ax^2)

Since the x and y are squared you will have two answers for each.
Without knowing the constants this is about as good as I can see to get it.

EDIT: Changed to new question.

ax + y = b ... (1)
ax^2 + y^2 = b^2 ... (2)

ax = b - y ... (1a)
(ax)^2 + ay^2 = ab^2 ... (2) * a

Substitute (1a) into (2) * a:

(b - y)^2 + ay^2 = ab^2