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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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?

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