## JoannaBlackwelder one year ago show that the normal line at any point on the circle x^2+y^2=r^2 passes through the origin

1. JoannaBlackwelder

@freckles

2. freckles

have you found y'

3. JoannaBlackwelder

Yeah, it is -x/y

4. imqwerty

this is a circle with the origin at center so any line tangent makes an angle of 90 with centr|dw:1443674243272:dw|

5. JoannaBlackwelder

Yeah, I got that too, but I thought that it needed to be more "calculusy" and less conceptual

6. JoannaBlackwelder

So, that is an option if I can't think of another way, thanks @imqwerty

7. imqwerty

ok lets think that way :)

8. freckles

$(a,b) \text{ is on the circle means } a^2+b^2=r^2 \\ \text{ so } b^2=r^2-a^2 \\ b=\pm \sqrt{r^2-a^2} \text{ assume } b>0 \text{ so we have } (a,b)=(a,\sqrt{r^2-a^2}) \\ \text{ the tangent line at } (a,b) \text{ is } \\ y-\sqrt{r^2-a^2}=- \frac{a}{\sqrt{r^2-a^2}} (x-a) \\ y=\frac{-a}{\sqrt{r^2-a^2}}(x-a)+\sqrt{r^2-a^2} \\ \text{ now the line that is perpendicular \to this one going through } (a,\sqrt{r^2-a^2} ) \\ \text{ is } y-\sqrt{r^2-a^2}=\frac{\sqrt{r^2-a^2}}{a}(x-a) \\ y=\frac{\sqrt{r^2-a^2}}{a}(x-a)+\sqrt{r^2-a^2} \\ \text{ so we need to show } (0,0) \text{ is on this line }$ just plug in (0,0) in the normal line there to verify then you can also do the same for the other assumption when b<0

9. freckles

and also when b=0

10. JoannaBlackwelder

Wow, thanks!

11. freckles

notice I didn't need to actually find the tangent line I just needed the slope of the tangent line and the slope of the normal line would be the opposite reciprocal of that

12. freckles

and then I was about to find the equation of the normal line from there

13. freckles

though we didn't really need calculus for this just algebra

14. freckles

so I kind of like @imqwerty 's way :) less work

15. imqwerty

(: thanks

16. freckles

|dw:1443675019274:dw|

17. freckles

that was just using the one algebraic formula for finding slope

18. imqwerty

umm m thinking of something like partial derivative of the equation of circle x^2+y^2 :/ what will it give?

19. JoannaBlackwelder

I found another way. :-) Since we got the slope of the tangent line to be -x/y, the slope of the normal line is y/x. Using the point (a.b), the new slope is b/a|dw:1443676370689:dw|

20. JoannaBlackwelder

So, the equation of the line has the y intercept as 0

21. JoannaBlackwelder

Which means the lines go through (0,0)

22. IrishBoy123

partials give you something like this write the circle as level curve: $$\phi = x^2 + y^2 = const$$ $$\large d \phi = \frac{\partial \phi}{\partial x} \ dx + \frac{\partial \phi}{\partial y} \ dy = <\frac{\partial \phi}{\partial x} , \frac{\partial \phi}{\partial y} >\bullet<dx,dy> = 0$$ because $$d (const.) = 0$$ but this also $$\implies \vec n_{x,y} = <\frac{\partial \phi}{\partial x} , \frac{\partial \phi}{\partial y} > \equiv<x,y>$$ for point P on circle at generalised point (a,b), $$\vec{OP} = <a,b>$$, $$\vec {n_P} = <a,b>$$ ie $$\vec{OP} \parallel \vec{n_P}$$ so $$\vec {n_P}$$ goes through O. which is all kinda like implicit diff in reverse because $$2yy' + 2x = 0, \; y' = -\frac{x}{y}, \; \vec t = <-y,x> \implies \vec n = <x,y>$$ as $$\vec n \bullet \vec t = 0$$

23. imqwerty

thanks ^_^ @IrishBoy123 :)