Find equations for the tangent line and normal line to the circle at each given point. (The normal line at a point is perpendicular to the tangent line at the point.)

Question

x² + y² = 25

(3, -4), (-4, -3)

At (3, -4)

Tangent Line: y=?

Normal Line: y=?

At (-4, -3)

Tangent Line: y=?

Normal Line: y=?

Answer 1

x^2 + y^2 = 25 is a circle with radius = 5

2x + 2yy' = 0

y' = -x/y = slope of a tangent line to the circle at point (x,y)

for (3,-4) slope = 3/4

for (-4,-3) slope = -4/3

tangent line through (3,-4) is y = 3x/4 + b where b = the y intercept

plug in (3,-4) to get -4 = 9/4 +b, b = -25/4

tangent line is y = 3x/4 - 25/4

normal line has negative inverse slope = -4/3

y = -4x/3 + b

plug in the point to solve for b

1. -4 = -4 + b
2. b= 0
3. y = -4x/3 is the normal line through the point

tangent line through (-4,-3) is y = -4x/3 + b

plug in (-4,-3) to solve for b:

-3 =16/3 + b

b = -25/3

tangent line is y = -4x/3 -25/3

normal line has the negative inverse slope 3/4

y = 3x/4 + b

plug in the point (-4,-3) to solve for b

-3 = -3 + b, b= 0

normal line is y = 3x/4

Answer 2

All of these are looking at the bottom half of the circle, so y = -sqrt(25-x²)

In order to find the tangent:

1) take the derivative and plug in at the point. The equation of the tangent at point (a,b) =

y = b + f'(a)(x-a) which is the formula for the line through point (a,b) and having slope f'(a).

Here the derivative is obtained by chain rule: dy/dx = f'(x) = -1/2 (25-x²)^-1/2(2x) = -x/(25-x²)^1/2

2) For the normal line, the slope will be the negative reciprocal of the slope of the tangent or -1/f(a) and, again, you can put that in point-slope form.