anonymous
  • anonymous
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!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Mertsj
  • Mertsj
If you take the first derivative of that expression you wll get (x-3)/(-x^2 +6x+16)^.5
Mertsj
  • Mertsj
That is the slope function. The slope at x = 0 is -3/4 found by replacing x with 0.
anonymous
  • anonymous
the first derivative is the difference quotient of the function right?

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Mertsj
  • Mertsj
So you have not yet had the rules for differentiation?
anonymous
  • anonymous
nope this problem was given to us on day one of class.
Mertsj
  • Mertsj
If not, here is another way. The radius is perpendicular to the tangent at the point of tangency as you learned in geometry.
anonymous
  • anonymous
and unfortunately my professor last semester never even mentioned the word limit or derivative....
Mertsj
  • Mertsj
Complete the square twice to find that the center of the circle is (3,4) and the radius is 5. Can you do that?
anonymous
  • anonymous
yes that was the only thing I was able to do and I got those same values
Mertsj
  • Mertsj
Since 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
  • anonymous
oh excellent! that was exactly what I needed!
Mertsj
  • Mertsj
|dw:1326904638924:dw|
anonymous
  • anonymous
so the slope of the tangent would be -3/4
Mertsj
  • Mertsj
. The slope at x = 0 is -3/4 found by replacing x with 0.
anonymous
  • anonymous
Thank 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!
Mertsj
  • Mertsj
you're welcome

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