Write the equation of the circle having a diameter with endpoints (-2,1) and (6,7)??
401455
Sorry.  This is going to be a little rough, typing the answer.  Here it goes:  \n(x-2)2(squared)+ (y-4)2 (squared)= 25\n\nYou find the midpoint between the two points for the center.  You find the distance between the two points for the diameter and divide that in half for the radius which equals 5.  The center is (2,4).  You square the radius and that is what comes after the equals sign.\n\nThe guy who entered before me is wrong.  I promise.
350225
C=2*π*r\n\nC-circumference\nr-radius\n\nd=7=2*r\n\nr=7/2\n\nC=2*π*(7/2)\n\nC=7π=7*3.14=22
844246
Okay, to begin with, rewrite the equation as so:\n\n(x/5)^2 + ((y+5)/6)^2 = 1\n\nso, if y = 0, the x = +/-sqrt(275/36)\n\nif x = 0, y = +1 or -11\n\nSo, the ellipse is symmetrical about the y axis, with the major axis endpoints at (0,1) and (0,-11)\n\nThe minor axis will be halfway between the major axis endpoints, so it will be when y = -5.\n\nWhen y = -5, x = +/-5\n\nSo, the minor axis endpoints are at (-5,-5) and (5,-5)\n\nNow, for the second part:\n\nFor an ellipse, let's say the distance between the foci is d. So, the distance from the furthest focus to the opposite major axis endpoint is d+L, where L is the distance from the nearest focus to the major axis endpoint. At the end of the minor axis endpoint, the distance from each focus is p. Now, 2p + d is equal to 2d+2L. We know that the entire length of the long axis is 11, and that is d+2L.\n\nSo, if 2d+2L = 2p+d, then d+2L = 2p\n\nd+2L = 11, so 2p = 11, or p=5.5\n\nNow, the right angle triangle formed by the endpoint of the minor axis, the intersection of the minor axis and the y axis, and one focus has sides 5 on one side, 5.5 on the hypotenuse and d/2 on the other side. Using Pythagoras' theorem to solve for d, we get:\n\n25+d^2/4 = 30.25\n\nso, d^2 = 21, or d=sqrt(21)\n\nSo, the focal points are at (0,-2.71) and (0, -7.29)\n\nqed
1010902
This is  basic\nslope= m=(y2-y1)/(x2-x1)=(12-7)/(8-6)=5/2\nangle=arctan(5/2)
862969
pi*(radius)^2*height\n\npi is aprox. 3.14 or 22/7
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
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
1119524
well using the y = function, you may need to create a positive and negative graph\n\nfor a circle with a center of 0, 0:\n\nsquare root ( x^2 - r^2)  (replace r with your desired radius)\nand\n- square root( x^2 - r^2)
39475
Diameter of a circle, I'm assuming?\n\nDiameter = 2 * Radius\nDiameter = Circumference / (2 * pi)\nArea = pi * (diameter / 2)^2
604208
The conjugate axis is your n term, sorta.  The general form of a hyperbola where the axes are on the x and y axes is\n\nx^2/a^2 - y^2/b^2 = 1\n\nSo if your conjugate axis is 6, then n = sqrt(6).\n\nThe foci are on the circle given by\n\nx^2 + y^2 = a^2 + b^2\n\nThis is really a circle.  Since your question says that the foci are sqrt(13) from the center, we can find a:\n\n13 = a^2 + 6\n7 = a^2\na = sqrt(7)\n\nSo your hyperbola has the equation\n\nx^2/7 - y^2/6 = 1
949680
The center is (-4, 6) and radius sqrt(64) = 8\n\nThis is because the equation of a circle with center (a,b) and radius r is\n\n(x-a)^2 + (y-b)^2 = r^2
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
799288
must be perpindicular to BC\nslope of BC is (0-(-1)/(6-1)=1/5\nslope of altityde is -5\ny=-5x+b & it imcludes point 3,2\n2=-5*3+b\n2=-15+b\nb=17\nthe equation is y=-5c+17
183956
Half the diameter, or circumference divided by 2.pi
55111
The formula for a midpoint is (X1+X2)/2 for X and (Y1+Y2)/2 for Y.  For your specific problem, the answer would be: (6,9)
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.
914363
Find the center-radious form of the equation of the circle with center = (2,-6) and the tangent to the y-axis.\n\nIf tangent to the y - axis then the point (0, -6) is the point of contact (as tangent to circle _|_ radius at point of contact)\n\nTherefore radius = 2\n\nTherefore required circle is\n\n(x - 2)² + (y + 6)² = 2²