Thread: Determining the intercept of a circle

1. Determining the intercept of a circle

Working in the first quadrant only.

I have a circle defined by the standard and known equation of
y = (r^2 - x^2)^-0.5

Where r is the constant radius of the circle.

y' = -x / (r^2 - x^2)^-0.5

0 < y < r
0 < x < r

I have a tangent line to the circle. I know where it intercepts the X axis at a value greater than d. It is the known variable.

I need to find a way to determine the slope of the tangent line or the (x,y) coordinate of where the line intercepts the circle. Actually solving one is solving the other.

intercept = function of distance

2. search derivative tangent slope.

https://gato-docs.its.txstate.edu/sl...t-Line/Finding

https://www.wyzant.com/resources/ans..._by_c_m_8_2_10

equation of a line y = mx +b were m is the slope dy/dx.

Derivative gives the slope at the point on the curve, dy/dx @ [x,y].

[x,y] is the point on the tangent line, m the slope, solve for b

b = y - mx

3. Thank you. I did find a solution.

Instead of using "y = mx + b" I used "y = m(x - c)" where c is the x intercept.

"y' = -x / (r^2 - x^2)^-0.5" means "m = -x / (r^2 - x^2)^-0.5"

Therefore by substitution I got

y = (-x / (r^2 - x^2)^-0.5)(x - c)

Setting that equal to the original equation for a circle I got

(r^2 - x^2)^0.5 = (-x / (r^2 - x^2)^0.5)(x - c)

Eventually by basic algebra I simplified that to

r^2 = x*c
x = r^2 / c

c is the intercept, so it is r + a, so we get

x = r^2 / (r + a)

Since r is a constant and a is the variable, I finally defined x in terms of a.

4. ok.

Plot a circle and the tangent line on top of it from your equations. You can probably do it in excel. See how it looks.

5. y^2 = [r^2 - x^2]^1/2

Applying the chain rule for u^n Du^n = n* [u]^n-1 * du

m = -x/sqrt(r^2-x^2)

At x = 0 m = 0 and you get a horizontal line tangent at y = r.

for r = 1 the midpoint in the 1st quadrant the point is (.707,.707).

m = -1 y = -x + 1.414

the x y intercepts are 1.414 and 1.414

m = -x/sqrt(r^2-x^2)
At x = r there is a zero. In the limit x goes to infinity the slope goes to infinity giving a vertical line tangent at x = r.

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