anonymous
  • anonymous
Need geometry help. The equation of a circle is (x + 6)2 + (y + 2)2 = 16. The point (-6, 2) is on the circle. What is the equation of the line that is tangent to the circle at (-6, 2)?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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.
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
I suppose you mean \((x+ 6)^2 + (y+2)^2 = 16\), correct?
anonymous
  • anonymous
yes
Jhannybean
  • Jhannybean
Wouldn't you have to take the derivative to find the slope, then plug the points (-6,2) and your slope into the form y= mx+b to find the y intercept? Then rewrite the equation for the tangent line?

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anonymous
  • anonymous
I hope you know where you're heading Jhannybean because this is not single variable analysis anymore.
Jhannybean
  • Jhannybean
Oh hahaha i was just guessing!! nevermind then :P
anonymous
  • anonymous
Line perpendicular to the line connecting the point and the center that also goes through the given point?
Jhannybean
  • Jhannybean
Yesss
anonymous
  • anonymous
The slope is zero at the point so it will just be the line y = 2. But this is only via visual inspection actually proving this is more difficult.
anonymous
  • anonymous
ok cool thanks for the help
Jhannybean
  • Jhannybean
|dw:1370068219672:dw| omg this is so not to scale :|
Jhannybean
  • Jhannybean
maybe that's how it's drawn? :\ not sure...
Jhannybean
  • Jhannybean
Probably not, haha, i tried!!
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=%28x%2B6%29%5E2+%2B+%28y%2B2%29%5E2+%3D+16%3B+y+%3D+2
anonymous
  • anonymous
thanks for trying
anonymous
  • anonymous
Visit the wolfram link for the graph he was trying to draw.
Jhannybean
  • Jhannybean
Ohhhhh I see it now!!!
anonymous
  • anonymous
thanks again for helping
anonymous
  • anonymous
I'm not sure how you can prove it formally without multivariable calculus.
Jhannybean
  • Jhannybean
Haha.i shall be learning that next semester! :P
anonymous
  • anonymous
Obviously, the picture is THE thing!
anonymous
  • anonymous
How to prove it with analysis? Notice that the circle is a manifold without boundary. We can define the chart \(\alpha(\theta) = (4\cos(\theta) - 6, 4\sin(\theta) - 2)\) \). Then just consider the tangent space of the chart at the angle corresponding to that point which is going to be \(\pi/2\)
anonymous
  • anonymous
Oh my!
Jhannybean
  • Jhannybean
@Alchemista are you done with all math?past multivariable calc? just wondering.
anonymous
  • anonymous
One is never done with all math. INFINITY DON'T YOU KNOW?
Jhannybean
  • Jhannybean
lol

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