Another quicker way is to impose that the circle slope is the same as the
line.
The first expression is the explicit equation for the top part of the
circle. Then I impose that the slope is that of the line and I find the
tangent `x` position.
y = y0 + sqrt(r**2 - (x-d)**2)
solve(diff(y,x) - m, x)
On Thursday, February 23, 2017 at 7:40:17 AM UTC+1, Michele Zaffalon wrote:
>
> 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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