More to my question... inscribed angle does it have to pass through the center of the circle?
401318
No, the only requirements for an inscribed circle are that the vertex has to be on the circle, and both of the rays have to intersect the circle.
86738
The angles all add up to 180 degrees.\n\nIf you know the lengths of all the sides you can use the "Law of Tangents" to find all the angles.  This is advanced geometry and Trigonometry, so rather than type, I am including a link to a pretty good explanation.\n\nhttp://answers.google.com/answers/threadview?id=38131
766220
If its distance from the center is less than the radius, it's inside.\n\nYou can find the distance from the center using your handy pythagorean theorem.
278253
5 lines?? must they be straight lines or can they be curved or another circle inside??  If you put the four lines thru like 2 x's then draw a small circle it is 16 pieces.
757335
x=r(t-sint) and y= r(1-cost) are the parametric equations of the cycloid where r is the radius of the rolling circle and t is the angle through which the radius of the circle has turned from its starting point. The angle t must be measured in radians.\n\nFor simplicity let r =1 so that we have the unit circle.\nwith t=60 = pi/3 we get\nx = pi/3-sin(pi/3) = pi/3 -(sqrt 3)/2\ny = 1-cos (pi/3) = 1- 1/2 = 1/2\n\ndy/dx= sint/(1-cost)\n=[(sqrt 3)/2]/(1/2) = 3^0.5 (at the point where t= 60= pi/3)\n\nFrom this , you can find the slope at any point define by the parameter t.\n\nb) The tangent is horizontal when t= pi and is perpendicular at the cusp which occurs at 2pi and multiples thereof. The maximum value  is 2r (or 2 for the unit circle). The length of the base of the arch is 2pi *r or just 2pi for the unit circle.
808242
Yes. These are intriguing geometry problems. What is YOUR question? Be specific. Post ONE question and be specific about where you are stuck. Right now, this just looks like you want someone to DO your homework for you. But you wouldn't cheat on your homework like that, would you?\n\nHere's a hint -- supplemental angles add up to 180\n\nFor example, if one angle is 60 degrees, its supplement is 120
896034
There is a theorem (I think you remember it better). I just put it with my own words. Let points A, B, C & D belong to the same circle. Then the angle ABC=(Arc ADC)/2\nThus the Sum of 7 Arcs under 7 Angles will be = to 2pi radians, that means that the Sum of those Angles = pi radians, the main demand being they belong to the circle.
831941
You're supposed to print a table, right? The question is poorly written, so it's not clear, but I'll assume that's the goal.\n\nWell, it sounds like you can write your loop, so shouldn't you just print out the sine, cosine, and tangent of the angle in the body of the loop?  It's not clear what the stumbling block is.  You realize that there are built in functions that compute sine, cosine, etc.?
747405
15 inches\nThe side of a regular hexagon inscribed in a circle is the same as the radius of the circle. In fact, if you connect the center of the circle to each vertex of the hexagon (with a straight line segment), then you will have 6 equilateral triangles that all meet at the center of the circle.
148902
its the middle of a angle like if da two lines combine the pointwere they combine is da vertex..........and you cant graph vertexs
380553
When we draw a circle with a compass, we keep the angle between the two legs constant.\n\nA ray from the top most point on the rainbow to our eye makes an angle to the horizontal direction. Theory says this angle is a constant.\n\nImagine the ray as one leg of long compass and the horizontal direction as another leg which points the center. Keeping the angle constant when the compass is rotated an arc is formed. \n\nIf we go high up we can see the whole circle.
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Hi. It is bent at any height above zero. Look at the horizon from a spacecraft and you'll see this. The bending gets reduced as you approach zero height, at which point it is a flat circle.
182934
You are given only Center (h,k) and radius r. Use the formula:\n(x - h)^2 + (y - k)^2 = r^2\nTransform this into:\nx^2 + y^2 + Ax + By + C = 0\nFor easier substitution.\n\nOr...\n\nWith the center (h,k) and radius r,\n\nPlot the points (h,k + r) (h,k - r) (h + r,k) and (h - r,k)\nThese points will be your guide to draw the circle!!
442031
If a rhombus is inscribed in a circle, then the rhombus is a square.  (Remember that a square is a rectangle with congruent sides, or a rhombus with congruent angles.)\n\nIf a trapezoid is inscribed in a circle, then the trapezoid is an isosceles trapezoid.
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The cosine laws apply to triangles which are NOT right angled. I think you have maybe read that wrong:\n\nhttp://www.themathpage.com/aTrig/law-of-sines.htm\n\ngives you a full guide to using the cosine laws to solve problems like these.
831242
This is just the inside of the unit circle including its circumference.\nThe unit circle (circle at the origin with radius=1) is\nx^2 + y^2 = 1\nAnd you want x^2 + y^2 ≤ 1\nso its the circle plus its inside.\n( if it had been x^2 + y^2 ≥ 1 it would be the circle plus everything outside )\nIf you wanted x in terms of y, you just get |x| ≤ √(1-y^2) .
263870
When we draw a circle with a compass, we keep the angle between the two legs constant.\n\nA  ray from the  top most point on the rainbow  to our eye makes an angle to the horizotal direction. Theory says this angle is a constant.\n\nImagine the ray as one leg of long compass and the horizontal direction as another leg which points the center.Keeping the angle constant when the compass is rotated an arc is formed.   \n\n Mathematically we say that the locus of all such points is an arc of a circle.\n\nIf water particle were present on all points on the arc we see a continuous arc of a rain bow.\n\nDuring cloudy  night we can see a full circle of rainbow \naround moon with moon as center.