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anonymous
 5 years ago
I am stuck on the 3rd part of a problem from the first assignment in calc 1! we were given the equation of a circle x^2+y^2−6x−8y=0
from which I took it to y=4 ± √(−x^2+6x+16)
we now need to find an equation of the tangent line to the circle at (0,0) my entire precalc course was essentially a diatribe on trig so I really need help in connecting these concepts! I am really hoping someone in here could walk through the process with me!
anonymous
 5 years ago
I am stuck on the 3rd part of a problem from the first assignment in calc 1! we were given the equation of a circle x^2+y^2−6x−8y=0 from which I took it to y=4 ± √(−x^2+6x+16) we now need to find an equation of the tangent line to the circle at (0,0) my entire precalc course was essentially a diatribe on trig so I really need help in connecting these concepts! I am really hoping someone in here could walk through the process with me!

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Mertsj
 5 years ago
Best ResponseYou've already chosen the best response.1If you take the first derivative of that expression you wll get (x3)/(x^2 +6x+16)^.5

Mertsj
 5 years ago
Best ResponseYou've already chosen the best response.1That is the slope function. The slope at x = 0 is 3/4 found by replacing x with 0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the first derivative is the difference quotient of the function right?

Mertsj
 5 years ago
Best ResponseYou've already chosen the best response.1So you have not yet had the rules for differentiation?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nope this problem was given to us on day one of class.

Mertsj
 5 years ago
Best ResponseYou've already chosen the best response.1If not, here is another way. The radius is perpendicular to the tangent at the point of tangency as you learned in geometry.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and unfortunately my professor last semester never even mentioned the word limit or derivative....

Mertsj
 5 years ago
Best ResponseYou've already chosen the best response.1Complete the square twice to find that the center of the circle is (3,4) and the radius is 5. Can you do that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes that was the only thing I was able to do and I got those same values

Mertsj
 5 years ago
Best ResponseYou've already chosen the best response.1Since the point of tangency is given to be (0,0) find the slope of the line between the center (3,4) and the point of tangency (0,0). The slope of the tangent will be the negative recipro cal of that slope.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh excellent! that was exactly what I needed!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so the slope of the tangent would be 3/4

Mertsj
 5 years ago
Best ResponseYou've already chosen the best response.1. The slope at x = 0 is 3/4 found by replacing x with 0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thank you! I really appreciate the help and that you stopped to help me look at the problem in a different light when I didn't see it from the first approach!
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