Implicit Differentiation Method

I get y' = -x/y as the derivative but can't figure out what to do next

1. Consider the circle x^2+y^2 = 1.
(a) At what point(s) is the slope of the tangent line equal to 1?
(b) At what point(s) is the slope of the tangent line equal to−1?
(c) At what point(s) is the slope of the tangent line equal to 0?

Derivative is correct.

(a)

Slope = 1
-x/y = 1
y = -x

x^2 + y^2 = 1
x^2 + (-x)^2 = 1
x^2 + x^2 = 1
2x^2 = 1
x^2 = 1/2
x = 1/√2, -1/√2

x = 1/√2 -----> y = -x = -1/√2
x = -1/√2 -----> y = -x = 1/√2

Derivative = 1 at points (1/√2, -1/√2) and (-1/√2, 1/√2)

(b)

Slope = -1
-x/y = -1
-x = -y
y = x

x^2 + y^2 = 1
x^2 + x^2 = 1
2x^2 = 1
x^2 = 1/2
x = 1/√2, -1/√2

x = 1/√2 -----> y = x = 1/√2
x = -1/√2 -----> y = x = -1/√2

Derivative = -1 at points (1/√2, 1/√2) and (-1/√2, -1/√2)

(c)

Slope = 0
-x/y = 0
x = 0

x^2 + y^2 = 1
0 + y^2 = 1
y^2 = 1
y = 1, -1

Derivative = 0 at points (0, 1) and (0, -1)

Mαthmφm

By definition, y' should be " - (y/x). The slope is 1 in the 2 nd and 4th quadrants and -1 in the 1st and
3rd quadrants.

By inspection, the slope is 0 at points {0,1} and {0, -1)