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I have to find the tangent lines to the circle x^2+y^2+6*x-8*y+25 = 1/16 which pass at the O(0;0) 

So i make a general line y=m * x

Ho can i put m*x instead of y in the circle and calculate the delta of the equation that i get?

 

I want to calculate the intersection between three circles.
I know that in this case i can calculate intersection of only the first and second equation, but I need this for a interactive component.

The command "intersection"[GEOMETRY] work only with 2 circles.

I did this but it doesn't work.

Thanks.

Parametric equation of a circle in 3d by three points. Draghilev method.

CIRCLE_3_POINTS_geom3d_2.mw

Hi.

In some cases when dealing with vectofields an such the are integral has to be expressed in terms of r(t).

the general form for r is r^2=(r*cos(t)-a)^2+(r*sin(t)-b)^2, When I solve this in maple it seems like I get the inverse of the desired result.

If I knew that was always the case I could just inverse my result to get the right expression for r, but im not sure if it only applies for this particular cas or all cases.

I would be happy if anyone took a quick look and suggested a way to obtain the desired solution for any center (a,b) for the circle.

 


 

Expression for radius, circle centred at (a,b)

RA := r^2 = (r*cos(t)-a)^2+(r*sin(t)-b)^2

r^2 = (r*cos(t)-a)^2+(r*sin(t)-b)^2

(1)

isolate(r^2 = (r*cos(t)-a)^2+(r*sin(t)-b)^2, r)

r = (1/2)*(a^2+b^2)/(cos(t)*a+b*sin(t))

(2)

eval(%, [a = -1, b = 0])

r = -(1/2)/cos(t)

(3)

plot3d([-2*x, x^2+y^2], y = -sqrt(-x^2-2*x) .. sqrt(-x^2-2*x), x = -2 .. 0, color = [green, red], orientation = [0, 0, 0])

 

(1/2)*Pi <= t and t <= 3*Pi*(1/2)

(1/2)*Pi <= t and t <= (3/2)*Pi

(4)

0 <= r and r <= -2*cos(t)

0 <= r and r <= -2*cos(t)

(5)

Area_off_center = int(r, [r = 0 .. -2*cos(t), t = (1/2)*Pi .. 3*Pi*(1/2)]); 1; Area_at_center = int(r, [r = 0 .. 1, t = 0 .. 2*Pi])

Area_off_center = Pi

 

Area_at_center = Pi

(6)

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Download parametrization_of_r_not_centred_at_orgin.mw
 

I would also happily like to know how I can solve for the range r can take, obviously in the example i´m working with here r starts at 0, but that is not always the case i guess.

 

Thank you, your help is much apperciated

how to Prove that the circumference of a circle of radius r is 2πr
on maple ?????

Suppose I have the parametric equations of a circle

x=cost

y=sint

where t runs from 0 to 2*pi. How can I show the orientation of this parametric curve on a plot?

Dear all,

I want to plot for example cos(theta) from 0 to 2*Pi inside a circle at a radius R. The axis theta of the ploted function is at a radius R.

Is it possible ?

Thanks

 

 

cilrcle.mw

i want to plot a circle which is centered at(0,0),and the radius is the length of Point2 and origin

but it shows some error,how could i do to solve this

 

cilrcle.mw

i want to plot a circle with that centered at (0,0),and the radius is the length of Point2 and orgin

but it shows the error

how could i do to solve this

 

 

 

 

 

 

 

 

c1 := circle([0, 0], 1, color = red);
p2 := implicitplot(x = 1/2, x = -2 .. 2, y = -1 .. 1.1, colour = blue, linestyle = 1, thickness = 2);
display(c1, p2);

How to fill that part beetween line and circle(where x>1/2 in circle )?

"Circular segment" is the unfortunate but standard term for the region between a chord and an arc of a circle sharing the same endpoints (see http://en.wikipedia.org/wiki/Circular_segment).  I say "unfortunate" because the phrase suggests a line segment when it actually means a planar region.

I would like to plot a shaded circular segment using Maple17.  I want the endpoints of the chord & arc to be anything I please, so the chord is not necessarily horizontal, or vertical, or the diameter of the circle, etc.

At the URL

http://www.mapleprimes.com/questions/139057-Segment-Of-A-Circle

there is an image containing a shaded circular segment, but I don't see what code produced the image.  The image there includes a horizontal chord, and I don't know if the code used to produce that image can be adapted for chords that are not horizontal.

If I have to, I can plot a shaded polygon with a huge number of sides that is indistinguishable from a circular segment.  I have plotted polygons before.  But it would obviously be preferable to plot a shaded circular segment.

If there a way to plot two curves of the form r = f(theta) and shade the region between them?  This would be better than the huge-polygon approach, but not as good as a simple command for plotting a shaded circular segment, if such a command exists.

In 2D, graph a blue ellipse x(t)=3cos(t), y(t)=2sin(t) for 0≤t≤2∏. For t=0.5 graph a green tangent line to the ellipse and a red osculating circle. Also, give the curvature, the equation of the tangent line and the center of the osculating circle

I think i'm missing something easy her , but I am trying to plot a circle and cannot figure out

what I am doing wrong. The equation for the circle is:

x^2+(y-3)^2=25

I define the equation

c := x^2+(y-3)^2 = 25

Solve it for y so I should be able to plot it with plot(cplot); :

cplot:=solve(c, y)

But what I get as answer is:

3+sqrt(25-x^2), 3-sqrt(25-x^2)

The equation of the circle is part of a homework assignment,...

I can't solve probably very easy problem. How to plot a filled semicircle which is rotated across one axis by an angle alpha? I came to the solution which I don't consider as the best one (since e.g. it will not work for alpha=Pi/2*(odd integer)) and I believe someone of you will show me a better approach. Thank you in advance.

My solution:

alpha := (1/6)*Pi:
plot3d([x, y, y*tan(alpha)], x = -1 .. 1, y = 0 .. sqrt(1-x^2)*cos(alpha), axes = normal, labels = ["x", "y", "z"...

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