determin the radius of a circle with the centre (0,0) and the eaquation x^2+y^2=49 iam guesin' its seven 7
351138
You got it.\n\nThe equation of a circle in standard cartesian from is:\n\n(x-a)^2+(y-b)^2=r^2\n\nThe circle is centered at (a,b) and r is the radius. For you, (a,b) is (0,0) and your r^2 term is 49, hence your r is the square root of 49 which is 7.
353472
2)(x-7)^2 + (y-6)^2 + (z-9)^2 = 36\n3)a)1 b) 3 c) 8\nFor 2 it is 36 because you use the absolute value of the smallest number of the center.  The smallest value is 6.\n6^2 = 36
515192
If it's a square, (60 x 60) then probably 8.\nIf it's a circle (r=60) then probably 10.\nIf it's a cirlce (d=60) then probably 6
312528
The radius of a circle is the distance from the center of the circle to the edge of the circle.
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
859740
This is not a rigorous way to do it, but it works all the same:\nYou know that 4=y^2+x^2 is the equation for a circle of radius 2 centered at the origin.  The line y=0 is just the x-axis.  So find the area of the top half:  A = ½πr² = ½π(4) = 2π\n\n*If you had actualy integrated it, it would have been ugly:\nint[√(4 - x²)*dx] from x=-2 to x=+2.
808986
y^2 = 1 - x^2\n\ny^2 + x^2 = 1\n\nremember that a circle centered on (x=0,y=0) has the property\n\nx^2+y^2 = r^2\n\nwhere r is the radius\n\nSo your plot is simply a circle arc with radius 1 centered on the origin.
981000
the equation for the circle is ( x-h)^2 + ( y-k)^2 = r^2\nnotice the plus sign (but  in your question there is a minus sign)\nso this equation is a hyperbola.
389938
The equation of a circle with the center at the origin is \nx^2 + y^2 = r^2 where "r" represents the radius of the circle\n\nTherefore the equation of a circle with the center at (0,0)and radius 10 units would be\n\nx^2 + y^2 = 10^2\n==> x^2 + y^2 = 100\n\n\nhttp://www.schoolingkids.com/\n\nhttp://www.futureaccountant.com/
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
510334
x = a*sin(ang)\ny = b*cos(ang)\n\nang goes from 0 to 360 degress, a and b determine how oval it is (if a=b it is a circle)
804813
(y+1)(y-7)\n=y^2-6y-7
53953
Area of a circle is pi times radius square.\n\nPi is 22/7 and radius is 2 squared\nA=22/7*2^2\nA=22/7*4\nAis apprx. 12.57
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.
292717
CIRCLE CHARACTERISTICS:\na) the equation contains the variable x & y.\nb) the highest degree for x is 2; the highest degree...\n...for y is 2.\nc) the coefficients of x^2 and y^2 are the same.\nd) the operation between x^2 and y^2 is plus (+),...\n...provided the sign before the x^2 is positive (+).\n\nSTANDARD EQUATION OF A CIRCLE:\n\n(x-h)^2 + (y-k)^2 = r^2 \nwhere:\nh = x-coordinate of the center of the circle.\nk = y-coordinate of the center of the circle.\nr = radius of the circle.\n\nEXAMPLE: \nProve that 4x^2 + 4y^2 - 24x + 32y - 44 = 0 is an equation of a circle.\n\nPROOF:\n\n4x^2 + 4y^2 - 24x + 32y - 44 = 0 ....................Given\n4x^2 - 24x + 4y^2 + 32y - 44 = 0 ...............Commutative\nx^2 - 6x + y^2 + 8y - 11 = 0....................Divide by 4\n(x^2 - 6x) + (y^2 + 8y) - 11 = 0................Associative\n{x^2 - 6x + 9) + (y^2 +8y +16) - 11 - 9 - 16 = 0  ......... \n..............Completing the Square\n(x^2 - 6x + 9) + (y^2 + 8y + 16) - 36 = 0 .....Subtraction\n(x^2 - 6x + 9) + (y^2 + 8y + 16) = 36 .......Transposition\n(x - 3)^2 + (y + 4)^2 = 36 ........Trinomial Perfect Square\n(x - 3)^2 + [y -(-4)]^2 = 6^2 ...Std. Equation of Circle\nwhere:\nh = 3; k = -4; r = 6\nMeaning: The above equation is an equation of a circle with center at (3,-4) and radius, r = 6.\n\nTherefore, since...\n4x^2 + 4y^2 - 24x + 32y - 44 = 0 can be transformed into a standard equation of a circle & having the characteristics of a circle, then:\n\nConclusion:\nThe equation 4x^2 + 4y^2 - 24x + 32y - 44 = 0 must be that of a circle.
392511
Find the distance between the center and the point (2,3).  This will be your radius.\nThen plug everything into the formula:\n(x-h)^2 + (y-k)^2 = r^2, where (h,k) is the center and r is the radius.
149999
A regular n-gon has the centers of its inscribed and circumscribed circles the same point. If you draw segments from this center to each of the vertex, you will get n number of radii and n number of triangles, which are all congruent. Using derivation you will arrive at the conclusion that\nA = .5aP or A = .5ans where\nA is the area, a is the apothem, s is the side and P is the perimeter\na =2A/ns\nThe radius is acquired by\nsin [180degrees(n - 2)/n] = a/r\nr = a/sin[180degrees(n - 2)/n]