This is how I would do analytically: the equation for the circle is (x-d)^2
+ (y-y0)^2 = r^2, a circle of radius `r`, centered at `(d,y0)`. You impose
that `y` that belongs to the circle, belongs to the line as well. To do
this, you substitute the equation for the line in that for the circle and
you obtain `expr` below (y -> mx).
If you solve this second order equation, you find two solutions because in
general, the line crosses the circle in two points (or never). You want
these two points to be coincident because this is an alternative definition
of tangent and you solve for `y0`. There are again two solutions, one with
the circle to the right of the line, one to the left. You should pick the
correct one: on my computer, this is the second solution.
If you now substitute this into any of `sols`, you will find your tangent
`x` point.
Does this help?
x, y, d, y0 = symbols("x, y, d, y0", real=True)
m, r = symbols("m, r", positive=True)
expr = (x+d)**2 + (m*x-y0)**2 - r**2
sols = solve(expr, x)
y0_sols = solve(sols[0] - sols[1], y0)
simplify(sols[1].subs(y0, y0_sol[1]))
On Thursday, February 23, 2017 at 6:54:12 AM UTC+1, Kevin Pauba wrote:
>
> I'm trying to wrap my brain around sympy and am looking for a little help.
>
> Let's say I have a line 'y=m*x' (y-intercept is zero). Let's also say I
> have a circle of radius 'r' whose center is constrained to be along a
> horizontal line a specific distance 'd' below the x-axis.
> Given m, r and d, how might I code sympy to determine the circle's x-value
> where the given line is tangent to the circle (i'm only interested in the
> tangent on the left side of the circle). I also would like to determine
> the (x,y) of the tangent point.
>
> Feel free to nudge me in the right direction if you don't have the time to
> offer a more complete solution.
>
> For those interested, my interest is related to the design a cam profile
> that uses a roller-type follower.
>
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