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Is there any way to draw the Radius "R4" with only the information supplied

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This is an unusual question for the Microsoft Access topic area. Did you intend to place the question under a different topic, or are you using MS Access/VBA to do this?

Rotate your figure clockwise until the line between the center of a circle and upper right point of your figure become vertical.

Imagine a set of circles with different sizes falling down to the angle between sides of your figure. Any circle will have 2 contact points, so there is no solution without additional data.

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(Cut an edge off a square post-it note, similar to your picture, and place a mug or other round-based object on it with the center at various distances from the sides to see the dilemma.)

We know the radius of the circle (R4)... but the original picture doesn't explicitly show the relationship between the circle and square/trapezoid (whether the circle is entirely contained, what portion of the arc intersects, etc).

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This is part of a drawing requirement and need to be done in access. As any of the values can vary so I need a formula. The numbers shown would always be known.

If we could somehow calculate the length of the line indicated we would know the center of the radius.

Are arcs drawn point to point or by angle?

How do you get the bisector of the angle without knowing the angle? The angle could be calculated?

So what is the minimum extra info that we need.?

I guess the angle of the slanted line in relation to the circle or vertical line might help?

There must be some trigonometry that gives the angle based on the two right angled lines?

Pi - ARCTAN(22 / 5)

>> So what is the minimum extra info that we need.?

If you make a complete circle, is your arc guaranteed to intersect each line in exactly/only ONE place (ie: are the lines tangential to the arc)?

If so, then I think you can use trig formulas to derive the distances from the corner to the intersect points using the previously derived angle and the radius. (Not exactly sure how, but I can dig into it if you don't get other responses)

If not, then you'd need some other piece of information to define the circle's position.

can anyone derive the center of the arc I assume I could then draw the arc using mbizup Pi - ARCTAN(22 / 5) . That's if arc are drawn that way. I have only drawn complete circles before but know that there are extra values that can be added after this:

Me.Circle (1000, 2000), (500), 255, ?,?

- draw perpendicular lines from the points where the 2 sides touch the arc
- at the point where these perpendicular lines intersect, this is the center of your circle which will give you an arc that touches both sides
- using a compass, draw the circle with a radius of either of the red radii (both are the same length)

Me.Circle (x, y), (Rad), Colour, ?,?

As you can see there are two other parameters that can be set. I have checked help in access 2002 but it certainly is not clear

https://msdn.microsoft.com/en-us/library/office/aa221156(v=office.11).aspx

other parameters are:

start angle = 0 (never tried, but I hope angles are started from 12), in article above –.00000001 is recommended

end angle = 3.14159265359 - ATN(22 / 5)

Nearly have it all done thanks to your help. Will post final solution when I have solved another question that I have just posted. It effects the physical drawing that access does so cannot confirm if formula is correct. To do with scaling. It draws circles as circles regardless of scaling but does scale lines and boxes.

Dim W As Variant

Dim H As Variant

Dim R As Variant

Dim A As Variant

Dim PI As Variant

Dim C As Variant

Dim Text As Variant

Dim LL As Variant

W = 20

H = 22

R = 4

A = 5 'Top right corner to intersection of where slanted line would meet to horizontal line

PI = 3.141592654

Me.Circle (W - A - (R * H) / Sqr(A * A + H * H) + A * R / H * (1 - A / (Sqr(A * A + H * H))), R), R, 0, Atn(A / H), PI / 2

Me.Line (W, H)-(W - A + A / H * (R - (A * R) / (Sqr(A * A + H * H))), R - A * R / (Sqr(A * A + H * H))), 0

Me.Line (W, H)-(0, H)

Me.Line (0, H)-(0, 0)

Me.Line (0, 0)-(W - A - (R * H) / Sqr(A * A + H * H) + A * R / H * (1 - A / (Sqr(A * A + H * H))), 0)

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PI = 3.14159265359

start angle = PI / 2 - Atn(22 / 5)

end angle = PI / 2