How to draw a circle with a dot?
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Put the pen on the paper and draw the center. Now fold one of the corners of the paper and bring the corner next to the pen. Now start moving the pen on the the this folded corner. ( Since the corner is folded you are actually not drawing anything on the front side of the paper. Now move as much as as the required radius on this folded part and then shift the tip of the pen back to the front side again. Now unfold the corner and since now you are back to the front part of the paper (but away from the center) complete the circle without leaving a trace of your shifting from the center to the circle.
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Draw a circle.  From the centre point to the edge draw a straight line.  This is the radius.  The area of a circle is pi times (radius squared.)  The Perimetre of a circle is 2 times pi times radius.
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Draw 2 concentric cirles 1 smaller and 1 larger then a teeth on the outer circle. Then use the circular array tool to array 6 the teeths or tooths (iuno) around the outer circle. And make it look like a gear.
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You are graphing y < -4/5  x + 9/5\n\nOn the y-axis, above zero, graph 1  4/5 (a little less than 2).  From that point, move down 4 squares and right 5 squares and put another point.  Connect this point with the 1 4/5 on the y-axis and draw a dotted line between them.   Shade BELOW  this line.
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since your ddoing coordinate geometry you should at least tell what shape you're trying to get.\nI'm assuming it's a parallelogram so u should make it (1,8)
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(m)\n\nBY  FOUR-CENTER  METHOD The   four-center   method   is   used   for   small ellipses.  Given  major  axis,  AB,  and  minor  axis, CD, mutually perpendicular at their midpoint, O, as shown in figure 4-45, draw AD, connecting the end points of the two axes. With the dividers set to  DO,  measure  DO  along  AO  and  reset  the dividers on the remaining distance to O. With the difference  of  semiaxes  thus  set  on  the  dividers, mark  off  DE  equal  to  AO  minus  DO.  Draw perpendicular  bisector  AE,  and  extend  it  to intersect the major axis at K and the minor axis extended  at  H.  With  the  dividers,  mark  off  OM equal  to  OK,  and  OL  equal  to  OH.  With  H  as a  center  and  radius  R1 equal  to  HD,  draw  the bottom  arc.  With  L  as  a  center  and  the  same radius as R1, draw the top arc. With M as a center and the radius R2 equal to MB draw the end arc. With K as a center and the same radius, R2, draw the  end  arc.  The  four  circular  arcs  thus  drawn meet,  in  common  points  of  tangency,  P,  at  the ends  of  their  radii  in  their  lines  of  centers
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Try it out in the lewis-dot diagram\n      ..        ..\nH-O-Ca-O-H\n      ..        ..
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When I was first trying to do trigonometry problems like this, I found it helpful to draw a picture.\n\nDraw a horizontal line to represent the ground, and then put a vertical line to represent the radio tower.  We want to know the height of the tower, so label it "x".  The angle between the tower and the ground is 90 degrees, so label it too.\n\nNow draw a dot (or a little stick figure with a protracter if you feel like it) to the side of the tower, on the ground.  Label the line between the tower base and the dot "50", as it's 50 meters from the tower.\n\nNow draw a line between the dot and the top of the tower, representing the person sighting to the top and measuring the angle.  Lable the angle between this sight line and the tower top "67 degrees".\n\nYou now have a drawing of the triangle with two angles and one adjacent side known.  This should be enough to get you started picking the right trig function to apply here.\n\nIf that's not enough to get going, here's another hint: Remember the unit circle with a triangle and the sides labeled with "sin(theta)", "tan(theta)" etc?  Are there any similarities to your tower drawing?  I always found being able to draw that unit circle on my scratch paper at the beginning of a problem set or test helped immensely.\n\nGood luck.
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http://local.wasp.uwa.edu.au/~pbourke/geometry/2circle/\n\nit tells you how
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Draw a horizontal line with a circle the size of your desired pentagon that has its center on this line. \nPut your compass' needle where the circle's circumference crosses the horizontal line, and draw a half-circle through the center of your first circle, crossing the circumference of the first circle in two places. Draw a vertical line through the points where the half-circle crosses the first circle. This line will pass through a point we call (a). \nOpen your compass so that you can, when placing the needle in the two intersections between the horizontal line and the first circle, draw a small cross above and below the horizontal line, outside the first circle, with one line of the cross from each point. If you join these crosses you will obtain a line perpendicular to the horizontal line, also passing through the center of the first circle. The point where this line crosses the circumference of the first circle on the top, we call (b). This is the first corner of the pentagon. \nPut the compass' needle in (a) and drawing a circle segment passing through (b) and down through the horizontal line, obtaining a point on this line we call (c). \nPut the needle in (b) and pass a circle segment through (c) and the first circle. These points on the first circle are the second and third corners of the pentagon. \nWithout extending the compass, put its needle in the second and third corners, and draw circle segments passing through the first circle to find the two remaining corners. Extend a line from each corner of the pentagon through the center of the circle made in step 1 of constructing a pentagon to the opposite side of that same circle. \nThe five corners of the pentagon constitute every other corner of the decagon. The remaining five corners of the decagon are those points where the lines of step 2 cross the original circle (but not a pentagon corner).
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You'll have to create it yourself. Here's how, with free, legal software:\n\n1. Download Inkscape, a free vector drawing program.\n2. Run it.\n3. File Menu -> Document Preferences\n4. Set Canvas Size to your paper size (Probably US Letter in the US or A4 elsewhere)\n5. Set Default Units to mm.\n6. Close dialog\n7. Create an ellipse with the ellipse tool\n8. Select the ellipse with the selection tool\n9. Set its width to 150 mm and its height to 86.1 mm using the text entries at the top of the window.\n10. Adjust the fill and stroke of the ellipse to whatever you like\n11. Print it.
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The following link  has a tutorial \n\nhttp://www.ncsu.edu/project/graphicscourse/gc/acadtut/acadtut2000/acadtut-home.html
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Yes, it is possible.\n\nDenote the number of black dots belonging to a red dot by N.\n\nThe present problem is formulated for N=10.\n\nLet us first solve a simpler problem with N=2.\n\nStep 1. \nTake a square, that is a four-sided regular polygon, with the side equal to R. Put the two red dots at the opposite [i.e., connected by a diagonal] vertices of the square. Put the two black dots at the remaining two vertices. If you draw a circle of radius R around each of the two red dots, you will find exactly 2 black dots lying precisely on the circle line. The problem is almost solved, but the number of red dots (2) is equal to the number of black dots (2).\n\nStep 2. \nTake another square and repeat the above procedure. \nNow we have the two square compounds. In total we have 4 red dots and 4 black dots.\n\nStep 3. \nPut the 5th red dot anywhere in the plane and draw a circle of radius R around it. \nDrag each of the above compounds as a whole closer to this circle. \nPlace one of the black dots belonging to the first compound and one of the black dots belonging to the second compound on the circle line surrounding the 5th red dot.\n\nWhile doing so, make sure that you move and rotate each compound as a whole, keeping all the sides of the squares equal to R. \n\nThe problem with N=2 has been solved: in total we have 5 red dots and 4 black dots satisfying the condition of the problem, 5>4. \n\nLet us solve the same problem for N=3.\n\nIn this case, one has to extend the Step 1.\n\nStep 1a. \nHaving constructed the square compound according to the Step 1 above, draw parallel vectors of length R starting at the vertices of the square. Make sure that the vectors do NOT coincide with the sides of the square. \nIf the vector starts at the red vertex, put the black dot at the end of the vector. If the vector starts at the black vertex, put the red dot at the end of the vector. \n\nNow we have one compound consisting of the two shifted squares containing 4 red dots and 4 black dots. If you draw a circle of R around each of the four red dots, you will find exactly 3 black dots lying precisely on the circle line.\n\nStep 2. Repeat the procedure 1+1a.\nStep 2a. Repeat the procedure 1+1a.\n\nNow that we have 3 compounds with equal numbers of red and black dots (in total 12 red dots and 12 black dots), put the 13th red dot anywhere in the plane and follow the instructions from Step 3 above.\n\nThe problem with N=3 has been solved: in total we have 13 red dots and 12 black dots satisfying the condition of the problem, 13>12. \n\nFor N=4, take the compound constructed for N=3 at Step 1a and translate ALL 8 vertices by a vector of length R.\nAgain, if the vector starts at the red vertex, put the black dot at the end of the vector and vice versa. You will have 8 red dots and 8 black dots, etc.\n\nThe above procedure applies to an arbitrary N, in particular to N=10.\n\n\nAn interesting question is: what is the minimal number of dots needed to solve the problem for a given N.
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well from my understanding there is 360 degrees in the circumference of a circle so divide that by 12 and you get 30 so every 30 degrees mark it.\nI hope i am not to wrong here!\n\nUnless its bigger but you can draw it down to fit that scale, that's not hard.
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I asked this a couple weeks ago:\nTo answer your question:\nVector graphics are composed of curves connected by points, mathematically defined. A circle is four curves connected by four points. A rectangle is four curves, connected by four points. (No, that wasn't a typo...they are considered curves even though they're straight lines. No, I don't know why...math and I never really got along all that well.) When you want to make the circle larger or smaller, the mathematical formula is redefined, and the circle keeps it's perfect crisp lines. Creature House Expression, Macromedia Freehand, and CorelDRAW are three examples of vector programs. These programs are considered object oriented because everything you make, even text, is an object that can be manipulated independent of other objects and the background (which can also be an object if you put something there, such as a rectangle covering the page.) You can arrange them so they are in front of each other, in back of each other, stack them, or group them. Vector programs are also sometimes called drawing programs, while programs like Adobe Photoshop and Procreate's Painter are paint, or raster programs.\n\nA raster program constructs what you see on screen by using pixels. When you enlarge the picture, the only way it gets bigger is for the pixels to get bigger too, and you lose the nice edges. So when you work in a raster program your best bet is to make your picture the size at which you'll ultimately be using it, because you will be limited in how much sizing you can actually do. And that is the beauty of working with vectors! You can work on a 4" object and enlarge it to 4' if you want. And it'll look just as good at 4' as it did at 4".\n\nSou
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Any circle that is tangent to two adjacent sides of a polygon has to have it's center on the bisector of the angle between them. Construct the bisectors of two of the angles and put the center where the lines cross.\n\nIf you construct a line from the center that is perpendicular to any of the sides, that gives you the radius of the circle.
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http://www.100hgh.com\n\nFold one side of the yarn down so that it is lined up in the center of the circle (refer to photo). This is in preparation to pull this yarn through the center of the circle.  \n\n\n Using your fingers or a crochet hook, pull the yarn through the center of the circle (first loop made).  \n\n\n If you haven't already, insert crochet hook into the loop just made.  \n\n\n Draw the loop up tighter by pulling with your fingers on the strands of yarn on both sides of the loop. Don't pull it too tight.  \n\n\n Step 6: Yarn over the hook.  \n\n\n Step 7: Pull this yarn through the loop on the hook (chain made). Hint Hold the last chain made between two fingers to keep it from drawing up smaller while you are pulling the next chain through. If you make your chains too tight, you'll find that you'll have difficulty in working the next row into these chains. If you keep your chains loose enough, crocheting the next row will be more of a pleasure! That's a good thing.  \n\n\n Repeat steps 6 and 7 for each chain that you need to make until you have made the number of chains needed in the starting chain for the pattern you are making.