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)?
I suppose you mean \((x+ 6)^2 + (y+2)^2 = 16\), correct?
yes
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?
I hope you know where you're heading Jhannybean because this is not single variable analysis anymore.
Oh hahaha i was just guessing!! nevermind then :P
Line perpendicular to the line connecting the point and the center that also goes through the given point?
Yesss
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.
ok cool thanks for the help
|dw:1370068219672:dw| omg this is so not to scale :|
maybe that's how it's drawn? :\ not sure...
Probably not, haha, i tried!!
http://www.wolframalpha.com/input/?i=%28x%2B6%29%5E2+%2B+%28y%2B2%29%5E2+%3D+16%3B+y+%3D+2
thanks for trying
Visit the wolfram link for the graph he was trying to draw.
Ohhhhh I see it now!!!
thanks again for helping
I'm not sure how you can prove it formally without multivariable calculus.
Haha.i shall be learning that next semester! :P
Obviously, the picture is THE thing!
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 \(\pi/2\)
Oh my!
@Alchemista are you done with all math?past multivariable calc? just wondering.
One is never done with all math. INFINITY DON'T YOU KNOW?
lol
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