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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Calculus1
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

have you found y'
Yeah, it is -x/y

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

this is a circle with the origin at center so any line tangent makes an angle of 90 with centr|dw:1443674243272:dw|
Yeah, I got that too, but I thought that it needed to be more "calculusy" and less conceptual
So, that is an option if I can't think of another way, thanks @imqwerty
ok lets think that way :)
\[(a,b) \text{ is on the circle means } a^2+b^2=r^2 \\ \text{ so } b^2=r^2-a^2 \\ b=\pm \sqrt{r^2-a^2} \text{ assume } b>0 \text{ so we have } (a,b)=(a,\sqrt{r^2-a^2}) \\ \text{ the tangent line at } (a,b) \text{ is } \\ y-\sqrt{r^2-a^2}=- \frac{a}{\sqrt{r^2-a^2}} (x-a) \\ y=\frac{-a}{\sqrt{r^2-a^2}}(x-a)+\sqrt{r^2-a^2} \\ \text{ now the line that is perpendicular \to this one going through } (a,\sqrt{r^2-a^2} ) \\ \text{ is } y-\sqrt{r^2-a^2}=\frac{\sqrt{r^2-a^2}}{a}(x-a) \\ y=\frac{\sqrt{r^2-a^2}}{a}(x-a)+\sqrt{r^2-a^2} \\ \text{ so we need to show } (0,0) \text{ is on this line }\] 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
and also when b=0
Wow, thanks!
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
and then I was about to find the equation of the normal line from there
though we didn't really need calculus for this just algebra
so I kind of like @imqwerty 's way :) less work
(: thanks
|dw:1443675019274:dw|
that was just using the one algebraic formula for finding slope
umm m thinking of something like partial derivative of the equation of circle x^2+y^2 :/ what will it give?
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|
So, the equation of the line has the y intercept as 0
Which means the lines go through (0,0)
partials give you something like this write the circle as level curve: \(\phi = x^2 + y^2 = const\) \(\large d \phi = \frac{\partial \phi}{\partial x} \ dx + \frac{\partial \phi}{\partial y} \ dy = <\frac{\partial \phi}{\partial x} , \frac{\partial \phi}{\partial y} >\bullet = 0\) because \(d (const.) = 0\) but this also \(\implies \vec n_{x,y} = <\frac{\partial \phi}{\partial x} , \frac{\partial \phi}{\partial y} > \equiv\) for point P on circle at generalised point (a,b), \(\vec{OP} = \), \(\vec {n_P} = \) ie \(\vec{OP} \parallel \vec{n_P}\) so \(\vec {n_P}\) goes through O. which is all kinda like implicit diff in reverse because \(2yy' + 2x = 0, \; y' = -\frac{x}{y}, \; \vec t = <-y,x> \implies \vec n = \) as \(\vec n \bullet \vec t = 0\)
thanks ^_^ @IrishBoy123 :)

Not the answer you are looking for?

Search for more explanations.

Ask your own question