0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.