Let's say that one has two fixed points with an arc of a specific radius drawn between them. How would one go about graphing this arc assuming that the center of the circle is not accessible. I have a feeling calculus would be involved? Is there a shortcut method that would provide a decent rough approximation giving a number of points along the radius to connect?

If the arc has been drawn then there's a center and a radius of curvature. What do you mean by "graphing" the arc? Also, any arc of a unit circle has a length the same as the angle it subtends, and there's a mathematical relation between the straight line (the chord) and the arc connecting the two points.

You mean like an arch structure above a door or gate? Can you post a picture of the problem you have in mind?

Looking back at my post, I can see the confusion. Let me try again. Ok. What I have basically is radius protruding from a wall. So, let's say we have a measurement along the wall from the corner to one end of the radius being one foot. Then, 6 feet from that point to the other end of the radius. Now, I need to draw an arc with a radius of 20 feet between those two points. I obviously cannot simply measure to the center of the circle as a wall is in my way. I need to use the relationship between the line and the radius to plot enough points to give me as accurate a radius as I care to achieve. The more points, the more accurate. How would I go about doing this? Arfa Explain, please. Exactly. Now, what is that relationship? That's the question. Wynn, Child's play. Layouts like that can generally be done in a large enough area to simply measure from the center of the circle to draw the radius. I could also draw a template of a given radius to solve the problem I have now, however I'd prefer to not have to go to such lengths if I could simply use a little math to plot a few points and connect the dots. As to a picture, if there is still confusion, I could cobble something together... I don't think I have any graphic software installed right now.

Just make a simple drawing by hand, mark what is known and what isn't or cannot be reached, and post the picture. This is geometry. Most people do geometry best by pictures ...

Ok. For a unit circle the length of a semicircle is pi. The length is the same as the angle, and this relation is always true for any arc or its chord. But that's for the largest chord possible, the diameter, and you want an arc of a circle with 20ft radius that has a chord 6ft long?

http://www.handymath.com/cgi-bin/arc18.cgi?submit=Entry http://www.1728.org/circsect.htm http://planetcalc.com/1421/

If I understand the problem, it is an easy one. The center of the circle in on the perpendicular bisector of the chord defined by the arc. Since you know the radius, you can find the center easily: From the intersection of the bisector & the arc, it is the distance equal to the radius.

If I understood what you're trying to do . . . can't you make a template by cutting a hole in some suitable material - a hole of radius equal to the radius of the arc - then cut a pie section out of your template, subtending the same angle as the one at your doorway? The edges of the pie section could be drawn before cutting the hole, and I think it would give a snug fit. If you want to approximate the arc by successive line segments (much harder but doable) you can treat the circle as a polygon of N sides, the solve for the interior angle between each side of the polygon, and then lay out the polygon using a protractor by successively laying out each interior angle. Maybe this will help: Please Register or Log in to view the hidden image!

Meh. The pair of points on the circle are also endpoints of a pair of radius lines, so you have a triangulation, an isosceles triangle with known sides. Use trig to get the angle at the center, the cosine rule or what have you. The length of the arc is equal to this angle when scaled by the actual radius. But maybe you want to know how the distance between the 6ft chord, along the wall, varies [from the arc itself] with the angle? Then using the successive approximation thing, as above, should let you "graph" it.

Using units of feet we have the following coordinates: corner = (0,0) point1 = (1,0) point2 = (7,0) radius = 20 center point = \((x_{\tiny c},y_{\tiny c})\) where \((x_{\tiny c} -1)^2 + (y_{\tiny c}-0)^2 = 20^2 = (x_{\tiny c}-7)^2 + (y_{\tiny c}-0)^2\) and \(y_{\tiny c} \lt 0\) so it follows that \((x_{\tiny c}-1)^2 = 20^2 - y_{\tiny c}^2 = (x_{\tiny c}-7)^2\) with solution of \((x_{\tiny c}-1)^2 = (x_{\tiny c}-7)^2\) being the solution to \(x_{\tiny c}^2 - 2 x_{\tiny c} + 1 = x_{\tiny c}^2 - 14 x_{\tiny c} + 49\) or \( 12 x_{\tiny c} = 48\) or \(x_{\tiny c} = 4\) and the solution to \((4-1)^2 = 20^2 - y_{\tiny c}^2\) being \(y_{\tiny c} = -\sqrt{400 - 9} = - \sqrt{391}\) So the equation for all points on the circle is \((x - x_{\tiny c})^2 + (y - y_{\tiny c})^2 = 20^2\) and \(y \gt 0\) or \(y = \sqrt{400 - (x - 4)^2} - \sqrt{391} = \sqrt{384 - x^2 + 8 x} - \sqrt{391} = \sqrt{(24 -x)(16 + x)} - \sqrt{391} \) The maximum distance it comes out from the wall is 4 feet from the corner where it comes out \(20 - \sqrt{391} \approx 0.22628\) feet or about 9/10 the way between \(2 \frac{11}{16}\) and \(2 \frac{23}{32}\) inches. The area of the chord of the circle that protrudes from the wall is \(\int_1^7 \sqrt{(24 -x)(16 + x)} - \sqrt{391} dx \approx 0.90614931\) square feet or a little less than 130.5 square inches. The circle chord can be approximated as the convex hull of many tangents. The following diagram exaggerates the y distance, but gives some examples of equations for tangent lines. http://www.wolframalpha.com/input/?...qrt[391], 3(7-x)/Sqrt[391]} for x from 1 to 7