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)?
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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.
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I suppose you mean \((x+ 6)^2 + (y+2)^2 = 16\), correct?
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?
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 \(\pi/2\)
@Alchemista are you done with all math?past multivariable calc? just wondering.
One is never done with all math. INFINITY DON'T YOU KNOW?