equation for circle with center C(3,-2) and radius 4.?
965380
(x - 3)² + (y + 2)² = 4²\n(C): x² + y² - 6x + 4y - 3 = 0
1180944
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
72316
The area of circle is pi*r^2 (pi r square). Now let the increased radius be R and iniitial radius be r.\nArea after expansion  =pi*R^2 (pi R square)\nArea before expansion =pi*r^2 (pi r square)\nIncrease in area      =pi*R^2  - pi*r^2\n                      =pi(R^2 - r^2) \n                      =pi(R+r)(R-r)\n                      =3*(7+5)(7-5), taking pi=3 aprox.\n                      =3x12x2\n                      =72 approx.\n Since answer c alone is near to this value the answer is C.\n\nThough the steps given are more the problem is done mentaly and not by writing these steps. The steps in mental calculations are \n1. 7+5=12\n2. 7-5=2\n3. 3x12x2=72 and choose the answer.
1033875
What "ChinNa90" gave was a very general form of a conic section equation - it does not have to be only circle, but also a parobola or ellipse or hyperbola. \n\nThe standard and easily understandable form of a circle in 2D Cartesian (x and y) coordinates is given by:\n\n(x - a)^2 + (y - b)^2 = r^2\n\nwhere:\nPoint (a,b) is the Centre of the circle\nr is the Radius\n\nTherefore, if you wanted the equation of a circle whose centre is (3,4) and radius is 7 units, then it's equation would be:\n\n(x - 3)^2 + (y - 4)^2 = 7^2\n\nx^2 + y^2 - 6x - 8y + 25 = 49\n\nx^2 + y^2 - 6x - 8y - 24 = 0
710286
the antiderivative of 4 / (t^2 + 1)  IS \n4 arctan(t) + C`
502090
Centre is 1.5,2 and the Radius is 3
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) .
776765
x^2 + (y-5/4)^2 = 1\n\nFirst, the circle must be on the y-axis by symmetry as zanti3 noted.\nSo the equation is in the form x^2 + (y-a)^2 = 1\nTo find the points of intersection with the parabola, y=x^2,\ncombine the equations thusly: \ny + (y-a)^2 = 1\ny + y^2 - 2ay + a^2 = 1\ny^2 + (1-2a)y + (a^2-1) = 0\nor \ny = [2a - 1 ± √(5-4a)] / 2\n\nNow, there is the easy intuitive way to proceed, and the brute force method to proceed.\nThe easy intuitive way is to note that the two values of y represent the intersection points of the circle and parabola.  But by symmetry (and also that the circle is tangent, not twice intersecting for (positive) x), we know that these two y-values must be the same!\nTherefore, \n [2a - 1 + √(5-4a)] / 2 =  [2a - 1 - √(5-4a)] / 2\nor \n√(5-4a) = 0\n5 - 4a = 0\na = 5/4; thus x^2 + (y-5/4)^2 = 1 is the equation!!!\n{(√3/2, 3/4) and (-√3/2, 3/4) are the points of intersection. Note that these points satisfy both equations.}\nFor a graph, see:\nhttp://mudandmuck.com/str2/circle-parabola.JPG\nNote that the circle is indeed "inscribed".\n\nOne brute force method is to take the derivitive of both functions, set them equal to each other and then substitute\ny = [2a - 1 ± √(5-4a)] / 2 for y.  You will get the same answer,\nas follows:\n\n(1) 2x + 2yy' - 2ay' = 0\n2x + y'(2y - 2a) = 0\ny' = 2x/(2a - 2y)\nand\n(2) y' = 2x\nso at the intersection points,\n2x = 2x/(2a - 2y)\n2a - 2y = 1\ny = a - 1/2\n\ncombined with:\ny = [2a - 1 ± √(5-4a)] / 2\na - 1/2 = [2a - 1 ± √(5-4a)] / 2\n2a - 1 = 2a - 1 ± √(5-4a)\n± √(5-4a) = 0\n5 - 4a = 0\na = 5/4
577034
Since the radius is 44, you could go either 44 units left, right, up or down from the center and find a point on the circle itself.  So (44,0) would be one such point.  However, these points are not in a quadrant; they are on an axis.  All the other points would fit in the circle's equation, x^2 + y^2 = 44^2.  So pick a number for the x-coordinate, square it, subtract this from 44 squared, then do the square root.  That gives you the y-coordinate.
1095202
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
181035
a circle's circumference is define as 2Pi x Radius\n\nin this case 2 x 3.14... x 3 so your answer is c
901448
Here's how to solve the problem:\n\nFirst find the center of the circle (call it (x1,y1).\n1. Write an expression for the distance of (x1,y1) from the first point, (1,8):  (x1-1)^2 + (y1-8)^2\n\n2. Write an expression for the distance of (x1,y1) from the second point, (-2,-1). (You should know how to do this, especially after seeing the example above.)\n\n3. Write an expression for the distance of (x1,y1) from the given line. This is a little trickier. There are probably various ways of approaching it. Here's one:\n\nRewrite the equation in slope-intercept form: y = (3/4)x +4\nThis tells us that the slope of the line is 3/4. So the slope of a line perpendicular to this line is -4/3. (The product of the two slopes has to be -1.)\n\nThe distance from (x1,y1) to the original line is measured along the direction perpendicular to that line. So it will be measured along a line with a slope of -4/3. The general equation for such a line is \ny = -(4/3)x + b, where b is the y-intercept of the line.\n\nNow we're trying to find an expression for the distance from (x1,y1) to the original line. To do this, we have to find where the perpendicular line intersects the original line. So we need to find the solution of the two equations, \ny = (3/4)x +4 and y = -(4/3)x + b.\nThis gives us two expressions for y, so we can set them equal:\n(3/4)x +4 = -(4/3)x + b\nSolving this, x = (12/25)(b-4) = (12b-48)/25\nAnd y = 9/25(b-4) + 4 = (9b +64)/25\nSo the point of intersection of the given line and the perpendicular line is ((12b-48)/25,(9b +64)/25).\n\nThe distance from (x1,y1) to this point (and to the given line) is \nsqrt((x1-(12b-48/25))^2 + (y1-(9b+64)/25)^2).\n\nHow do we find b and x1 and y1? We need to equate the three expressions for distance (this one, and the two found in items 1 and 2 above). Solving these equations will give us the center of the circle and the equation for the perpendicular to the given line.\n\nNext find the radius of the circle:\nDetermine the distance from the center to either of the two given points (using the Pythagorean theorem ... that is, using the distance formula). Call this distance D.\n\nFinally, write the equation of the circle:\n(x-x1)^2 + (y-y1)^2 = D^2
74729
The formula for the area of a circle is:\nPi*r^2/2\nto translate, "one half of Pi times the suare of the radius"\n\nSo, if you have a 3 cm radius:\n(3 cm)(3 cm) = 9 cm^2\n\nthen divide by two:\n9 cm^2 /2 = 4.5 cm^2\n\nthen multiply by Pi:\n\n4.5(Pi) cm^2\n\nIf you want a decimal number, you can use an estimation of Pi, something like 3.1415, but it's not exact since Pi is an irrational number (a non-ending and non-repeating decimal)
344528
The formula is as follows:\n\nC\n- = pi\nd\n\nC=the circumference\nd=the diameter\nand pi=3.14\n\nThe next step is to put all the known information into the equation:\n\nC\n- = 3.14\n14\n\nOnce you have that, you can multiply 14 by both sides to get C alone. Like this:\n\n14( C/14) = (3.14)14\n\n14 is canceled on the side with C, and the you take 3.14 and multiply it by 14 to get 43.96. So now the equation looks like this:\n\nC = 43.96\n\nAnd there you have your answer. The circumference of the circle with a diameter of 14 is 43.96\n\nHope this helped.
549743
Area increases 4 times. Beacause Area of circle = PiR2.\n\nSo if R doubles then radius will be 2R.\n\nHence area will be Pi(2R)2. which results into\n\n4 Pi R2.
1167722
areah of a circle=pie times r squared  (r is radius)\ndiameter= 2r\nso r= d/2\n\nso you do 6sqrt(3)  divided by 2\n(you get 3 times sqrt of 3)\nthen do that number sqared....times pie(3.14)\n(18 times pie...more accurate if use calculaterbut answer is 18x3.14......)
841586
(0,5/4)\n\nI solved this problem a few days ago using algebra.\n\nYou take the derivative of x^2, take the negative reciprocal (it will be -1/(2x)  This will be the equation of the line you are looking for.  The tangent points of the circle and parabola is perpendicular to the radius which goes through the center.  You have two different formulae for the slope, and the fact that the radius is one (use distance formula).  Then solve it and get the answer.\n\n\n\nSorry that's not by Maple, but it can be used to double-check your answer I guess.  If you want me to work it out in full, add a note and I will.\n\n\nHopes this helps.