Find two point on a circle x^2+y^2=1 where the slope of the tangent line is -2 write the equation for the normal lines at these points.

To find two points on the circle where the slope of the tangent line is -2, we need to differentiate the equation of the circle with respect to x, get the slope using the derivatives, and then solve for the points.

1. Differentiation:
The equation of the circle is x^2 + y^2 = 1. Differentiate both sides of the equation with respect to x.
d/dx (x^2 + y^2) = d/dx (1)
2x + 2y(dy/dx) = 0

2. Get the slope:
We want to find the slope of the tangent line when dy/dx = -2. Therefore, substitute dy/dx = -2 into the equation obtained from the differentiation.
2x + 2y(-2) = 0
2x - 4y = 0

3. Solve for x and y:
Solve the above equation, 2x - 4y = 0, simultaneously with the equation of the circle, x^2 + y^2 = 1. There are several ways to solve this system of equations, including substitution or elimination. Here, we will use substitution.

From 2x - 4y = 0, we can solve for x:
2x = 4y
x = 2y

Substitute x = 2y into the equation of the circle:
(2y)^2 + y^2 = 1
4y^2 + y^2 = 1
5y^2 = 1
y^2 = 1/5
y = ±√(1/5)

Substitute the values of y back into the equation x = 2y:
If y = √(1/5):
x = 2(√(1/5)) = 2√(1/5) = √(4/5)

If y = -√(1/5):
x = 2(-√(1/5)) = -2√(1/5) = -√(4/5)

Therefore, the two points on the circle where the slope of the tangent line is -2 are (√(4/5), √(1/5)) and (-√(4/5), -√(1/5)).

4. Equation of the normal lines:
To find the equation for the normal lines at these points, we need to find the slope of the normal lines and then write the equations using the point-slope form.

The slope of the normal line is the negative reciprocal of the tangent line's slope since the normal line is perpendicular to the tangent line. Therefore, the slope of the normal line is 1/2.

Using the point-slope form of a line (y - y₁) = m(x - x₁), where (x₁, y₁) is a point on the line and m is the slope, we can write the equations for the normal lines:

For the point (√(4/5), √(1/5)):
(y - √(1/5)) = 1/2(x - √(4/5))

For the point (-√(4/5), -√(1/5)):
(y + √(1/5)) = 1/2(x + √(4/5))

These are the equations for the normal lines at the given points on the circle.