I know that 3 points define a circle and the point equidistant from all 3 is the center of that circle. Given the 3 points, how do I find the center? Is the equation different since we are working with earth coordinates?
137835
sphere points: A, B, C as vectors from sphere center O.\n\ncompute:\n\nX = A x B\nY = A x C\n\n\nU = X x (A + B)\nV = Y x (A + C)\n\ncenter points:\n\nP = ± U x V\n\n(if necessary normalize P to sphere radius)\n\n--------------\n\ncross product: A x B\n\nsketch:\nhttp://static.flickr.com/48/129027255_d097e01a2d_o.jpg\n\nfurther reading: spherical Voronoi Diagrams, projective geometry\n\nthanks Neodiogenes. my solution was incomplete. i corrected it.
477328
Here's the answer you want:  With a compass set to the radius of the circle, start marking off radii around the circle, dividing it into 6 equal parts.  Connect any 3 alternate points and you'll have your triangle.\n\nNO MATH REQUIRED!
640439
no, a circle has infinite lines of symetry. For an equilateral triangle, i believe there r only 3
487769
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.
17103
The third principle just needs to be adjusted algebraically for your situation. \n\nAssuming a circular orbit:\nF=GMm/r^(2+a) and F=ma, where a=r(w^2).\na=centripetal acceleration; w=angular velocity\n\nGMm/r^(2+a)= mr(w^2)\nNow substituting w=2pi/T where T is period of motion.\n\nGMm/((r^2)+(r^a))=\n4mr(pi^2)/T^2\n\nGM/((r^2)+(r^a))=\n4r(pi^2)/T^2\n\n((r^2)+(r^a))/GM=\n(T^2)/4r(pi^2)\n\nr((r^2)+(r^a))/GM=\n(T^2)/4(pi^2)\n\n((r^3)+(r^(a+1)))/GM=\n(T^2)/4(pi^2)\n\nT^2=\n[4(pi^2)/GM]*[(r^3)+(r^(a+1))]\n\nThis last line would be the answer you would get for Kepler's third principle with the your parameters. His original law is T^2=[4(pi^2)/GM]*(r^3). I dropped the negative sign in your equation just to make the math easier, so my equation represents the force on one body and yours with the negative would represent the force on the other body, equal in magnitude and opposite in direction. The periods, T, should still be the same.\n\nI don't believe the second principle would change at all, since it is just a statement of the conservation of angular momentum. I'm not sure about how these changes would affect the second principle, so you might want to check another source.
673075
alright, ill tell you, but not without explaining it.  im giving you the benefit of the doubt that you are not just cheating on your homework :)\n\nthe standard equation of a circle, using only variables is\n(x-h)^2 + (y-k)^2 = r^2\nwhere the center is at (h,k) and the radius is r\n\nthat means that when you look at the equation youre given, the x coordinate of the center is whatever number is subtracted from the x\n\n(x+1) ---> (x- -1)  so  h = -1\n\nyou use the same process to determine "k", which is the y coordinate of the center\n\n(y+3) ---> (y- -3)  so  k = -3\n\nthat makes the center of the circle at (-1, -3)\n\n\nfinding the radius is also simple using the standard equation. here it is again, just for reference:\n(x-h)^2 + (y-k)^2 = r^2\n\nsee how you can use that to get this?\n9/4 = r^2\n\nso just take the square root of both sides . . .\nr = +/- 3/2\n\nbecause you took the square root, the 3/2 could technically be either positive or negative, but since youre talking about the radius of a circle and you can't have a negative distance, you only take the positive answer.\n\nso r=3/2
560385
The first point is not at exactly 210 degrees.  It is a little over 212 degrees.\n\nThe second point (3,3) is obviously at 45 degrees, not 30 degrees.\n\nThe distance between (-8, -5) and (3, 3) is\nsqrt((-8 - 3)^2 + (-5 - 3)^2) = sqrt((-11)^2 + (-8)^2) = sqrt(121 + 64)\n = sqrt(185)\n\nThe original angle from (3,3) to (-8, -5) is\natan2(-8 - 3, -5 - 3) = atan2(-11, -8) = 180 degrees + arctan(8/11)\ncounterclockwise.\n\nThe clockwise rotation subtracts thirty degrees from that, so in the final coordinate system the original (-8, -5) is sqrt(185) magnitude in the 150 degrees + arctan(8/11) direction from the new origin at the original (3,3).\n\nLuckily, sin( arctan(8/11) ) = 8 / sqrt(8^2 + 11^2) = 8 / sqrt(185)\nand cos( arctan(8/11) ) = 11 / sqrt(185).\n\nWe have the "new" x is\n\nsqrt(185) cos(150deg. + arctan(8/11)) =\n = sqrt(185) ( cos( 150 deg) cos ( arctan(8/11) ) - sin( 150 deg ) sin( arctan(8/11) ) )\n = 11 cos(150 deg) - 8 sin(150 deg)\n = 11 (-sqrt(3)/2) - 8 (1/2)\n = -(1/2)(8 + 11 sqrt(3)) or approx. - 13.526\n\nand the "new" y is:\n\nsqrt(185) sin(150deg. + arctan(8/11)) =\n = sqrt(185) ( sin( 150 deg ) cos ( arctan(8/11) ) + cos( 150 deg ) sin( arctan(8/11) ) )\n = 11 sin( 150 deg ) + 8 cos( 150 deg )\n = 11 (1/2) + 8 (- sqrt(3) / 2)\n = (1/2)(11 - 8 sqrt(3)) or approx. -1.428\n\nwhich makes sense, as the original slightly over 212 degrees after the rotation is slightly over 212 - 30 = 182 degrees, so both quantities are negative with the magnitude of x large and the magnitude of y small.\n\nCheck: "new x"^2 + "new y"^2 =\n = ( -(1/2)(8 + 11 sqrt(3)) )^2 + ( (1/2)(11 - 8sqrt(3)) )^2\n = (1/4)( 8^2 + 2*8*11 sqrt(3) + 3*11^2 + 11^2 - 2*11*8 sqrt(3) + 3*8^2)\n = (1/4)( 4 (8^2 + 11^2) )\n = 185 as it should be.\n\nCheck: atan2("new x", "new y") = 180 degrees + arctan("new y" / "new x")\n\n"new y" / "new x" =\n = (11 - 8sqrt(3)) / (- 8 - 11sqrt(3))\n = (11 - 8sqrt(3)) (- 8 + 11sqrt(3)) / ( (-8)^2 - (11sqrt(3))^2 )\n = (-88 + 121sqrt(3) + 64sqrt(3) - 88*3) / (64 - 121*3)\n = (- 352 + 185sqrt(3)) / -299 or approx. 0.105587\nwith arctan 6.02737\nand expected value arctan(8/11) - 30 or approx. 6.02737\nso this checks out, too.\n\nChanged coordinates:\n\n( -(1/2)(8 + 11 sqrt(3)), (1/2)(11 - 8 sqrt(3)) ) or approx ( - 13.526, -1.428 )
391437
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.
531805
The Earth orbits the sun in an ellipse.\n\nEvery ellipse has two foci. It is at these two points where the orbited body is normally located.\n\nPlease see the wiki source that has further information on how to get the numerical data if you need it.
621127
The equation of a circle is:\n(x-h)^2 + (y-k)^2 = r^2\nwhere:\nh: x-coordinate of the center\nk: y-coordinate of the center\nr: Radius of the circle\n\nThe center of the circle will be midway between the points. On the x-axis:\n\n(-4 + 8)/2 = 4/2 = 2\n\nOn the y-axis:\n\n(2+8)/2 = 10/2 = 5\n\nSo the location of the center as coordinates is (2,5)\nThe radius is simply the distance from the center to either point. So, using the pathegreom theorm and the center at (h,k) and the point at (Px,Py):\n\nsqrt((h-Px)^2 + (k-Py)^2) = r\n\nUsing (8,8) for the point and the center at (2,5):\n\nsqrt((2-8)^2 + (5-8)^2) = sqrt((-6)^2 + (-3)^2) = sqrt(36 + 9) = sqrt(45)\n\nObviously, you could plug square root of 45 into your calculator to get some decimal equivilent, but it's not necessary since the equation requires r^2 anyways. So our final equation, using the form above, is:\n\n(x-2)^2 + (y-5)^2 = (sqrt(45))^2\n\nSimplifying, the square root and ^2 cancel out leaving:\n\n(x-2)^2 + (y-5)^2 = 45
433548
Since the foci are along the y-axis, the major axis of the ellipse will be along the y-axis.  Also, the center of the two foci is the center of the ellipse, which will be (0,0) or the origin.\n\nSo the only thing left to figure out is the length of the minor axis.\n\nThe major axis, minor axis, and focal length (distance from the foci to the center) are related by the Pythagorean theorem:\n\na² + b² = c²\n\nYou know that a = 26/2 = 13, c = 12, so b = 5.\n\nTherefore the equation is:\n\nx²/25 + y²/169 = 1
230452
Centre is 1.5,2 and the Radius is 3
549425
Orthocentre = point where altitudes intersect (An altitude is the line from the vertex to the opposite side (extended if necessary))\n\nIncentre = centre of the incircle = point where perpendicular bisectors of each of the sides intersect. \nRadius of incircle = √[(s-a)(s-b)(s-c)/s] where s = semiperimeter\n\nCentroid = point where medians meet (Median is the line joining a vertex to the midpoint of the opposite sides) The centroid is the centre of gravity of the triangle and the centroid divides each median in the ratio 2:1 On a numberplane the centroid of a triangle is the point which is at the average of the sum of the three x- and y-coordinates of its vertices)\n\nOne other is:\nCircumcentre = centre of the circumcircle = point where angular bisectors meet.\nDiameter of circumcircle = a/sinA = b /sinB = c/sinC
650281
Example - you'll need to apply it to your own endpoints.\n\nA circle is described on the line joining (-2,5)and(3,4) as diameter, find its equation:\n\nThe equation for a circle is:\n(x-h)^2 + (y-k)^2 = r^2 where (h,k) is the center and r is the radius\n\nIt is easiest to find the center first. it is the midpoint of the line and the slope is constant:\nh = x1-[(x1-x2)/2] = 3 - 5/2 = 1/2\nk = y1-[(y1-y2)/2] = 5 - 1/2 = 4.5\n\nNow, find the radius using the distance formula to find the distance between the center and the endpoint or by using a right triangle where dx and dy are the lengths of the legs. I prefer the latter, but they are pretty much the same:\ndx^2 + dy^2 = r^2\nr^2 = (5/2)^2 + (1/2)^2= 26/4 = 13/2\nBy substituing back into the equation for a circle:\n(x-.5)^2 + (y-4.5)^2 = (6.5)^2 = 42.25
491137
This type of problem will eventually seem EASY for you, so hang in there.  Let's begin by stressing that "a circle with radius 5" is simple to draw -- I recommend marking the 4 points that are "5 units" above, below, right and left of the center point.\n\nThat is, mark (-6,7) on your graph paper, then sketch four more points at (-1,7), (-11,7), (-6, 2) and (-6,12).  "Connect the dots" by drawing all 4 quarter-circle arcs between the points.\n\nThe equation for any circle -- in "rectangular coordinates" (which means (x,y)) -- is normally:  (x-a)^2 + (y-b)^2 = r^2, where (a,b) is the center point and r is the radius.\n\nBut what you're hoping to find is the equation of a circle in "polar coordinates" (which means (r,theta)).\n\nThe reason a sample triangle was drawn inside the given circle (I'm assuming you're looking at a textbook) is to illustrate the relationships among x, y, theta and r.  Let's inspect, using rules from trignometry:\n\nx = r cos (theta) + a\ny = r sin (theta) + b\nr = sqrt ((x-a)^2 + (y-b)^2)\ntheta = Arctan ((y-b)/(x-a))\n\nwhere (x,y) is any point on the circle, (a,b) is the circle's center, and r is the radius.\n\nThe formula for theta that I just provided may be precisely what you are looking for.  It works simply by first finding the "slope" of the line connection (x,y) and (a,b), then taking the inverse tangent of that slope, which 'converts' it to an "angle."\n\nHope this helps!
545963
Let E be the raduis of the earth\nLet r be the actual radius of the circle in the plane\nLet S be the radius of the circle on the sphere\nLet θ be the angular separation of the centre of the circle on the sphere to the circle in degrees\n\nThus θ = πθ/180 radians\n\nThen C = 2πr\nr = E*sinθ → C = 2πE*sinθ\nAlso S = E*πθ/180 → E = 180*S/(πθ)\n\nThus C = 2π*sinθ*180*S/(πθ)\nie C = 360Ssinθ/θ \nwhere S = radius on sphere and θ is the angular separation (at the centre of the sphere) between the centre of the circle and any point on its circumference, expressed in degrees
596859
the antipode.  if you draw a line from x, y through the exact center of the earth, the point x+180, -y will be where that line exits the earth's surface on the other side (assumes the earth is a perfect sphere, which it is not).\n\ngood luck!