## anonymous one year ago Last question! YAY! How can I show that the normal line at any point on the circle x^2 + y^2 = r^2 passes through the origin?

1. anonymous

first show that slope of tangent line is dy/dx $\frac{dy}{dx} = -\frac{x}{y}$ slope of normal is opp reciprocal $m = \frac{y}{x}$ next use equation of normal line to show y-intercept is 0 $y - y_0 = m(x - x_0)$ $y = mx + (y_0 - mx_0)$ $y = mx + (y_0 - \frac{y_0}{x_0} x_0) = mx$ Since y-intercept is 0, every normal line must go through origin

2. anonymous

Thank you! That makes sense now. :D

3. anonymous

yw :)

4. anonymous

Oops! One last thing. How did you get -x/y for the tangent line?

5. anonymous

@dumbcow

6. anonymous

taking the derivative of circle equation and solving for "dy/dx"

7. anonymous

Oh, sorry. Never mind. I think I figured it out.

8. anonymous

Thank you. :)