Find the radius of the circe with the equation x^2+y^2-6x-12y+36=0?
917697
The general form of the equation of a circle is:\n\n(x-k)^2 + (y-h)^2 = r^2\n\nwhere the ordered pair (k,h) represents the center of the circle and r = radius.\n\nTo get the x and y squared terms, you need to form a perfect square.\n\nFor the x terms, you have [x^2-6x].  You can deduce that you should get the perfect square term (x-3)^2 because when you FOIL this out, you get a -6x term.  (x-3)^2 = x^2-6x+9, thus you need a '9' term to complete the above perfect square.\n\nUsing the same logic for the y term, you have [y^2-12y], thus the perfect square you can deduce is (y-6)^2.  FOILing this out, you get y^2-12y+36, thus you need a '36' term to complete the perfect square.\n\nTo get these constant terms, we just need to rewrite the constant value in a different way.\n\nThus, you get\n\n(x^2-6x+9) + (y^2-12y+36) - 9 = 0\n\nIf you combine the constant terms in this equation, you still get 36 because 9+36-9 = 36. Now you can solve for the general form of a circle:\n\n(x-3)^2 + (y-6)^2 = 9\n\nUsing the general form of the circle equation, you can conclude that r^2 = 9.  Therefore, radius = 3\n\n-------------------\n\nHope this helps
796855
if you mean an ellipse....\n\narea is equal to pi * minor radius * major radius
563113
radius is half of the diameter.\n\nyou can use the circumference equation; c = pi x d\n\nd = 2r so then c = pi x 2r\nc / pi = 2r\nc / pi / 2 = r
986257
If the equation is X^4 - 12x^2 -64 =0, there is.\n\nUse u = x^2, now your equation will look like this:\n\nu^2 - 12u - 64 =0\nFind u1 and u2 using the quadratic formula.\n\nThe solutions are u1 = 16, u2 = -4\n\nNow you go back to x, first deal with u1, then u2.  Because u = x^2 each u solution will give you two x solutions.\n\n\nu = x^2 = 16\nx = +/- SQRT(16)\nx1 = 4\nx2 = -4\n\nx2 = -4.  This is not possible with real numbers. so it will give you two imaginary numbers. Do the same as before\n\nx = +/- SQRT(-4) = +/- 2*SQRT(-1) = +/- 2i\n\nx3 = i\nx4 = -i
860968
u can use the following formulas to find the radius\n1)Area of circle=pi.*(radius)^2\nwhen area is given plug in that value into this formula and u can get the radius\n2)radius=diameter/2\nwhen the diameter is given,find the radius by dividing the diameter by 2\n3)circumference=2*pi.*r\nwhen the circumference is given use the above formula to solve for the radius\n\nthere is no definite formula for the radius but depending the other value given u can solve for the radius.\nhope that helps!!!
237156
Pi * r^2\n\n(Pi R Squared)\n\nwhere R is the radius (which is also equivalent to half the diameter)
1160449
x^2 + 9y^2 - 24x + 18y + 9 = 0\nor (x^2 - 2*x*12 + 12^2) + 9(y^2 + 2*y*1 +1)  + 9 - 12^2 - 9 = 0\nor (x - 12)^2 + 9(y + 1)^2  = 12^2 = 144\nor (x - 12)^2/144 + 9(y + 1)^2/144 = 1\nor ((x - 12)^2)/12^2 + ((y + 1)^2)/4^2 = 1\n\nTransform the equation to new coordinate system\nX = x - 12 and Y = y + 1\n\nStandard equation of ellipse in the new coordinate system is\nX^2/a^2 + Y^2/b^2 = 1\n\nFor this equation, the centre is at (X=0,Y=0) in new coordinate system.\n i.e. (x - 12 = 0, y + 1 = 0) i.e. (12,-1) in old coordinate system\n\nThe coordinates of foci are (X = a*e, Y = 0) and (X = -a*e, Y = 0) in new coordinate system.\n\nQuestion: What is 'e'?\nAnswer: 'e' is the eccentricity of the ellipse and e = sqrt[1 - (b/a)^2]\nHere a = 12, b = 4\nSo e = sqrt[1 - (4/12)^2] = (2/3)*sqrt(2)\n\nSo in new coordinate system foci are (X = 8*sqrt(2), Y = 0) and (X = -8*sqrt(2), Y = 0)\n\nIn old coordinate system foci are (x = X + 12 = 12 + 8*sqrt(2), y = Y - 1 = 0 - 1 = -1) and (12 - 8*sqrt(2), -1)\n\nVertices are points where the ellipse cuts the major axis.\nThese are (X = a = 12, Y = 0) and (X = -a = -12, Y = 0) in new coordinate system.\nSo in old coordinate system the vertices are (24,-1) and (0,-1).\n\nI wrote vertex instead of centre in the previous solution. But I have corrected it now.
273731
120\n\n(6y-10)^2 = 36y^2 -60y -60y +100
4959
=8(y^3)z + 12(y^2)(z^2) - 20y^2
589996
Assuming that A, B, C are all on the same line.\n\nLarger circle:\nRadius = AC = 6\nArea = PI * r^2 = 36*PI\n\nSmaller circle:\nRadius = AC - AB = 6 - 2 = 4\nArea = PI * r^2 = 16*PI\n\nShaded area:\nArea = (Large A) - (Small A)\n     =  36*PI - 16*PI = 20*PI = 62.8 units^2
627033
6y + 3 = -4y^2\n4y^2 + 6y + 3 = 0\n\na=4, b=6, c=3\n\ny = -b+-SQRT(b2-4ac) / 2a\ny = -6+-SQRT[36-(4)(4)(3)] / [(2)(4)]\ny = -6+-SQRT[-12] / 8\n\nBut SQRT[-12] is not possible, unless we are looking for imaginary values, then:\n\ny = -6/8 +- SQRT[-12]/8\ny = -3/4 +- [2SQRT3]i/ 8\ny = -3/4 + [(SQRT3)/4]i or -3/4 - [(SQRT3)/4]i
439090
((x^2-2x)/4y)/((x^2-4)/6)\n=(6x(x-2))/(4y*(x-2)(x+2))\n=3x/(2y(x+2))
1167592
6√2\n\nRemember it's a square so the sides are all the same length, and the area is s^2\n\nso s^2=36==>s^2-36=(s-6)(s+6)=0 so s is either 6 or -6.  -6 makes no sense here, so it is obviously 6.\n\nNow, since we have a square we can "draw" (imagine) a line at the diagonal, and that makes a right triangle (since the corners of a square are all 90º) with legs 6 and 6.  using the Pythagorean relation, we have d^2=s^2+s^2=6^2+6^2=72.  Thus d^2-72=(d-6√2)(d+6√2)=0.  Thus d is either 6√2 or -6√2, and again is obviously the positive one :)
227387
D = 36 - (4)*(-2)*(-3)= 36- 24 = 12\nTwo real solutions.
642779
6^2 = 36, so 6^-2 = 1/36\n2^-3 = 1/8\n\nMethod:  Raise the number to the positive power, then use 1 as the numerator and the result as the denominator.\n\nNow, remember the method.
612346
Fact or out the x. This leaves 36x^2-84x+49 which is a perfect square (6x-7) so it facotrs to:\n\nx(6x-7)(6x-7)
178452
y^2 + 3y + 6 -4y^2 - 6y simplifies to:\n\n-3y^2 - 3y + 6\n\nIt's a second-degree trinomial expression.