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!
If you take the first derivative of that expression you wll get (x-3)/(-x^2 +6x+16)^.5
That is the slope function. The slope at x = 0 is -3/4 found by replacing x with 0.
the first derivative is the difference quotient of the function right?
So you have not yet had the rules for differentiation?
nope this problem was given to us on day one of class.
If not, here is another way. The radius is perpendicular to the tangent at the point of tangency as you learned in geometry.
and unfortunately my professor last semester never even mentioned the word limit or derivative....
Complete the square twice to find that the center of the circle is (3,4) and the radius is 5. Can you do that?
yes that was the only thing I was able to do and I got those same values
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.
oh excellent! that was exactly what I needed!
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