I am trying to find the $x$ coordinate on a circle which has a gradient of 8. I am trying to do this by taking the derivative of the circle equation and then using the derived equation to figure out the $x$ value for when the equation = 8. Below is the equation for the circle which I am working with.

Circle equation:
$\sqrt{50 - (x - 27.071)^2} + 45$

Derived equation where the gradient = 8:
$8 = -\frac{x - 27.071}{\sqrt{(50-(x - 27.071)^2}}$

The issue I am having is that I am finding it difficult to calculate $x$ in the derived equation. I am hoping that someone may be able to assist in solving the derived equation as I am struggling to do so and come up with the correct answer. Any help with the question will be very much appreciated, thank you.


So the equation to be solved is $$ 8 = -{x - b\over\sqrt{50 - (x - b)^2}}, \quad\mbox{ where } b=27.071.\tag{1} $$ Multiplying by the denominator and squaring we get $$ 64(50 - (x - b)^2) = (x - b)^2, $$ therefore we have a quadratic equation: $$ (x - b)^2 = {640\over13}, \quad\mbox{so}\quad x-b=\pm\sqrt{640\over13}, $$ and we find the solution $$ x=b-\sqrt{640\over13}=27.071-\sqrt{640\over13}\approx20.0545. $$ (The other root of the quadratic equation does not satisfy $(1)$, so we discard that root.)

Note: Logically, however, a full circle must have two points of equal slope 8 (these two points are the endpoints of the same diameter). We got only one point - because you actually started with an equation of the upper half-circle, rather than the full circle.

  • $\begingroup$ Great answer thanks for all your help. $\endgroup$ – Nick0014 Apr 16 '17 at 1:45

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