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Forums
Mathematics
Calculus
Minima/maxima of a circle
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[QUOTE="Mark44, post: 6862454, member: 147785"] Your first equation is of a circle whose center is at the point (a, a) and whose radius is r. A more general equation would put the center at an arbitrary point (a, b), say. I have a couple of problems with your second equation, though. First, this says that the derivative is zero. That's not true in general and is true only for two points: (a, a + r) and (a, a - r). Second, the derivative should have two values for a given value of x. Starting with the equation ##(x - a)^2 + (y - a)^2 = r^2##, implicit differentiation gives ##2(x - a) + 2(y - a)\frac{dy}{dx} = 0##. Solving for dy/dx gives $$\frac{dy}{dx} = \frac{a - x}{y - a} = \frac{a - x}{\pm\sqrt{r^2 - (x - a)^2}}$$ Note the plus/minus I've added. [/QUOTE]
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Forums
Mathematics
Calculus
Minima/maxima of a circle
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