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How Do I Find the Equation of the Tangent From Any Point Outside the Circle to That Circle?

Let M(p, q) be the point outside the circle.Let C(x1, y1) be the center of the circle and r it's radius. Let T(x, y) be the point of tangency. CT will be normal to TM. Eqn of TM : y- q = m (x - p). Slope of CT = -1/m. Eqn of CT: my x = my1 x1. Solve for intersection of CT and TM which is T(x, y) in terms of p, q and m.

1. Generating visually pleasing circle packs

The idea is toInstead of working with graphics primitives and the position of circles and their radii I work with the image matrix directly.Code:Adjustable parameters are the paddings, the size of the circles and for how long it continues to try to pack the shape.With the color scheme BlueGreenYellow and shape = DiskMatrix we getWith the color scheme DeepSeaColors andwe getFinally, this is a circle packed map of Sweden using the color scheme DarkTerrain:The careful observer will note that the smallest objects are not actually round. Do not worry about this, it's because the smallest object is just one pixel and you can not make a circle out of that. It saved me time to generate the graphics like this, it can easily be fixed by making a larger image and setting the smallest circle to a radius of say three or five, then shrinking the image to whatever size one wants

2. 3D coordinates of circle center given three point on the circle.

I think using a projection into 2D might be the easiest way to actually calculate. If the points are \$A, B, C\$ then find \$\$beginalign mathbfu_1 & = B-A mathbfw_1 &= (C-A) times mathbfu_1 mathbfu & = mathbfu_1 / | mathbfu_1 | mathbfw & = mathbfw_1 / | mathbfw_1 | mathbfv & = mathbfw times mathbfu endalign\$\$This gives three orthogonal unit vectors with \$mathbfu\$ and \$mathbfv\$ spanning the plane. Get 2D coordinated by taking the dot products of \$(B-A)\$ and \$(C-A)\$ with \$mathbfu\$ and \$mathbfv\$. Let \$\$beginalign b & = (b_x,0) = ( (B-A) cdot mathbfu , 0 ) c &= (c_x,c_y) = ( (C-A) cdot mathbfu, (C-A)cdot mathbfv ) endalign\$\$We know the center must lie on the line \$x= b_x/2\$. Let this point be \$(b_x/2,h)\$. The distance from c must be the same as the distance from the origin \$\$(c_x-b_x/2)^2 (c_y - h)^2 = (b_x/2)^2 h^2\$\$ So \$\$h = frac(c_x-b_x/2)^2 c_y^2 - (b_x/2)^2 2 c_y \$\$The actual center can then be recovered by taking \$A (b_x/2)mathbfu h mathbfv\$.This is a nice explicit calculation.

3. How to get a circle over A

Do not redefine

as this is the 'ring' used over ''

4. Determining the Diameter of a Circle

Yes is valid. I do not know where is your confusion but notice that \$x^2y^2geq 0\$. Therefore \$sqrtx^2y^2\$ is always well defined in the sense that you get a real number as a result, and of course is legal two divide by two. So the expression is mathematically valid and you are able to calculate it for any point \$(x,y)\$ of the space. The reasoning of deduction of the formula is also flawless.

5. Open sets on a circle

An open set is technically a closed set, taken out the boundary. In your case, for simplicity let us assume \$1\$ be the circumference of the circle. Then \$(0,1)\$, the set of all points from \$0\$ to \$1\$ excluded the two points at the ending is one such open set. Hope that helps :)

6. circle becomes an oval after saving

I just ran into the same issue editing a layer in WGS 1984. I created a new shapefile without a defined projection and the distortion did not occur after I saved the edits. Looks like Zbynek was correct, it is a projection issue. Try projecting the layer into what your mxd is projected in and maybe redrawing the circle

7. Is there some sort of device that let's you cut a near perfect circle/half circle?

For large circles, there are arrangements that rotate a sheet of metal under a laser or cutting torch

8. Intercept of sine with circle

We find maximum and minimum of functions:For sinusoide:\$y=4 sin (0.5 x -1) implies y' = 0.5 cos (0.5 x -1)= 0 implies x/2 -1= pi/2 implies x_s=pi 2\$ and \$y_s =4\$ is maximum point. Also sinusoide cross x axis at points \$2, 0\$ and \$(x = 2pi 1 7. 28, 0) \$ For circle maximum is \$(x_c, y_c)= (4, 5)\$ and minimum is \$(x_c, y_c)=(4, 1)\$We can see that \$2

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