what is the center of the circle & radius for.... x squared + y squared=25 ??  Answer?   (_,_) radius=___?
608708
The basic formula for a circle is (x-h)^2 + (y-k)^2 = r^2\n\n(h,k) would be the center (where x=h and where y=k), and the radius is the positive square root of r^2\n\nSo in your example, there is no h or k, so the center would be (0,0) and the radius would be the positive square root of 25, which is 5.
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
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.
456513
A=pi*x*y \n\nwhere x is small radius from centre to circumference and y is large radius from centre to circumference.
796855
if you mean an ellipse....\n\narea is equal to pi * minor radius * major radius
325867
First, I'll assume you mean that this is an ellipse (parenthesis are important):\n((y+2)^2)/25 + ((x-3)^2)/16 = 1\n\nTo graph by hand, start at the origin.  \n1)The center is shifted- to find it's location, set (y+2) = 0 and (x-3) = 0, and solve for x and y. The center is therefore at (-2,3). \n\n2)The half length of the axis (the lines bisecting the ellipse) are given by the denominators of the x and y terms.  To find the length of each axis, take the square root of each denominator.  The "radius" in the y (vertical) direction is sqrt(25) = 5 units long, and in the x direction it is sqrt(16) = 4 units long.  \n\n3)Starting at the center point of the ellipseee, (-2,3), count up half the length of the axis in the y direction, or 5 units, and put a point.  Do the same thing down, and then repeat with the x axis, counting left and right 4 units, and putting points on your graph.  Draw an ellips through these 4 points, and you're done! \n\nThe major axis is therefore the y axis, and is 10 units long, and then minor axis is the x axis, which is 8 units long.\n\nIf you want to graph this on a calculator, you must solve for y:\n((y+2)^2)/25 + ((x-3)^2)/16 -((x-3)^2)/16= 1 -((x-3)^2)/16\n\n(((y+2)^2)/25)*25 = (1 -((x-3)^2)/16)*25\n\n((y+2)^2)^(1/2) = ((1 -((x-3)^2)/16)*25)^(1/2)\n\nNote that raising to the (1/2) power is the same as taking the square root. When you take the square root of a real number, your answer could be negative or positive, so you have 2 solutions, one for the bottom of the ellips, and one for the top:\n\n(y+2) = (((1 -((x-3)^2)/16)*25)^(1/2))\nand\n(y+2) = -(((1 -((x-3)^2)/16)*25)^(1/2))\n\nThe final equations you'd enter into your calculator are:\n\ny = (((1 -((x-3)^2)/16)*25)^(1/2))-2\nand\ny = -(((1 -((x-3)^2)/16)*25)^(1/2))-2
625973
How to do this problem:\n1. Find the Radius. To do this, take the diameter and divide by two. This gives you a radius of 5.\n2. Because the Area of a circle is equal to Pi times the radius squared, and your teacher wants the answer in terms of pi, square the radius and get 25 pi.
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!!
771214
Must have x + y > 0, else the expression is negative or ±∞. Multiply by x + y to get\n\n¼ ( x + y ) < x² + y² < ½ ( x + y )\n\nx² + y² - ¼ x - ¼y > 0  and x² + y² - ½ x - ½y < 0\n\n( x - 1/8 )² + ( y - 1/8 )² > 1 / 32 and ( x - ¼ )² + ( y - ¼ )² < 1/8\n\n( x - 1/8 )² + ( y - 1/8 )² > 1 / 32  are all points outside a circle centred at ( 1/8 , 1/8 ) with radius √2 / 8.\n\n ( x - ¼ )² + ( y - ¼ )² < 1/8 are all points inside the circle centred at ( ¼ , ¼ ) with radius √2 / 4.\n\nThe first circle is inside the second and touches it at (0,0).\n\nThe required condition is satisfied by all points outside the small circle and inside the large circle. Use QuickMath (Plot -> Advanced) to see the region.
802152
Circumference\n\nC = 2 π r\n\nC = 2(3.141592654)(1)\n\nC = 6.283185307\n\nC = 6.28 Feet rounded to two decimal places\n\nThe answer is C - 6.28\n\n- - - - - - - - - - - - - - - -\n\nArea of a circle\n\nA = π r²\n\nA = 3.141592654 (1)²\n\nA = 3.14 Squared feet. Rounded to two decimal places\n\nThe answer is A - 3.14\n\n- - - - - - - s-
762369
Find the coordinates of the points where the straight line meets 'y=1' meets the circle 'x^2+y^2-14x+2y+45=0'.\n\n'x^2+1^2-14x+2(1)+45=0\nx^2 -14x +48 = 0\nthe solutions are:\nx=(14 +- sqrt( 14^2 -4(48) ) /2\nx=14+- sqrt(4) )/2\nx= (14 +2)/2 or x= (14-2)/2\nx=8 or x = 6\n.
510980
yeah, looks like the circle is just a 0, which would mean x^0 + 7 = 1 + 7 = 8.  It does look weird, though.
807243
I'm not sure how far I'll get with this one, but I'll give it a try.\n\nLet the center of the circle be at (p,0), so the equation of the circle is (x-p)^2 + y^2 = a^2, with a>p.  The chord is y = mx, and we want to find where the chord and circle intersect.\n\n(x-p)^2 + m^2 x^2 = a^2\nx^2 - 2px + p^2 + m^2 x^2 = a^2\n(m^2 + 1) x^2 - 2px = a^2 - p^2\n\nSolve by completing the square (details omitted) to get\n\nx = {p +/- sqrt[m^2 (a^2 - p^2) + a^2]} / (m^2 + 1)\n\nThe center of the chord (hence center of the circle with chord as diameter) is at x = p / (m^2 + 1), with y = mx = mp / (m^2 + 1).  The radius is sqrt[m^2 (a^2 - p^2) + a^2] / (m^2 + 1), so the equation of that circle is\n\n[x - p / (m^2 + 1)]^2 + [y - mp / (m^2 + 1)]^2\n  = [m^2 (a^2 - p^2) + a^2] / (m^2 + 1)^2\n\nIf q = m^2 + 1, then by clearing fractions, we get\n\n(qx - p)^2 + (qy - mp)^2 = m^2 (a^2 - p^2) + a^2\n\nI'm not going to check my work, but those are the answers I ended up with.
788538
x^2+2x+y-10y+22=0\nis NOT a circle, there is a y^2 missing.\nthis is a parabola!!!! '\n\nbut if you meant\nx^2+2x+y^2 -10y+22=0\nthen\nx^2 + 2x +1 -1 + y^2 -10 y + 25 -25 + 22 =0\n(x+1)^2 + (y-5)^2 =4\n\nso center (-1,5) \nradius: 2 \n'
71470
3.1416 X radius squared\n12" circle= 6" radius\n6" X 6" = ? X 3.1416(pi)= ?\nGot the picture?\nNot going to do your homework for you.\n\n(sorry Lucky Lady but pi X diameter is the circumference)
481336
well, ok, i see lots of complicated answers...but here is the answer in simple terms\n\nthe radius measures teh distance from the center to its boundry...it doesnt matter what the coordinates of teh boundry are because we have to use the distance formula to find it which is\n\nd = sqrt[(x-x1)^2 +(y-y1)^2]\n\nso from this equation (x1,y1) is the origin (0,0)\n\nso you really have\n\nradius = d = sqrt[x^2 + y^2]\n\nso whatever point you choose the distance must be positive because any numbers squared then added will be positive.\n\nso you would claim, well what if the center wasnt at the origin, well the only difference would be that you would use the original distance equation i mentioned.\n\nSo we can say that all distance (according to the distance equation) must be positive, try it with any points you like, it will always be positive.  Due to the distance equation always being positive, all radii must be positive.\n\nhope this helps\n\nmatttlocke\n\nPS\nand for everyone who says distance cant be negative...they lie\n\nthe magnitude of direction can never be negative the negative does imply direction.  for example, on the number line, we have points -5, and 5, the distance of both of those points from 0 is 5, but if we were plot each point they would be going in opposite directions, and we know this due to the negative sign\n\nso the distance formula means that the distance will always be positive, the negative does tell you direction, ask any mathematician (im working on that) or a physicist and they will tell you the same thing.\n\nBut, a negative and positive radius mean the same thing because the boundry of the circle will intersect all the points at whatever the measure of the radius is in all directions, so i spose you could look at it as negative, but it doesnt really matter.  This would all hold in any 'n' dimension.
419498
First, group the terrms with x in parenthesis and the terms with y in parenthesis:\n(x^2 + 6x)  + (y^2 - 8y) = -9\nLet's think for a moment, we have two squared terms, x^2 and y^2. Think of the two groups in parenthesis as if you were completing the square to solve a quadratic equation.\nFor both, take 1/2 the middle term and square it, adding it to both sides.:\n(x^2 + 6x +(3)^2) + (y^2 - 8y + (4)^2) = -9 + (3)^2 + (4)^2\nFactor and combine like terms:\n(x + 3)^2 + (y - 4)^2 = 16\nNow we have the standard form of a circle\nThe center is the OPPOSITE of the numbers in parenthesis, according to PLNR (positive left, negative right)\nThus the center is (-3, 4) and radius is sqrt(16)= plus/minus 4\nRemember that the radius must extend in negative directions: down and left. That's why it is plus/minus.