find the center-radious form of the equation of the circle with center = (2,-6) and the tangent to the y-axis.
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
1172110
let circle =  x^2 + y^2 = r\nin xyz plane.\n\n1 line is y = ax ,   a is any real no,\n2nd is z axis
226433
The circle's center is at (4, -1).  Therefore its radius is sqrt((6-4)^2 + (3- -1)^2) or sqrt 20.\n\nSo its equation is (x-4)^2 + (y+1)^2 = 20\n\nThe reason we know the center is at (4, -1) [at least how I did it] is by calling the center (a,b) and writing three different but equal applications of the distance formula, one from each known point to (a,b), then taking two pairs of these three distances, making them equal, and solving for a and b.
979191
When converting from rectangular coordinates to polar coordinated, remember the following.\n\n@ = Theta\nx = r*cos(@)\ny =r*sin(@)\n\nr^2 = x^2 +y^2\n\ntan(@) = y/x\n\nso your equation\nx^2 +y^2 =36 becomes\nr^2 = 36   OR\nr =6 (which is a circle with radius 6)
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
900535
This type of problem will eventually seem EASY for you, so hang in there.  Let's begin by stressing that "a circle with radius 5" is simple to draw -- I recommend marking the 4 points that are "5 units" above, below, right and left of the center point.\n\nThat is, mark (-6,7) on your graph paper, then sketch four more points at (-1,7), (-11,7), (-6, 2) and (-6,12).  "Connect the dots" by drawing all 4 quarter-circle arcs between the points.\n\nThe equation for any circle -- in "rectangular coordinates" (which means (x,y)) -- is normally:  (x-a)^2 + (y-b)^2 = r^2, where (a,b) is the center point and r is the radius.\n\nBut what you're hoping to find is the equation of a circle in "polar coordinates" (which means (r,theta)).\n\nThe reason a sample triangle was drawn inside the given circle (I'm assuming you're looking at a textbook) is to illustrate the relationships among x, y, theta and r.  Let's inspect, using rules from trignometry:\n\nx = r cos (theta) + a\ny = r sin (theta) + b\nr = sqrt ((x-a)^2 + (y-b)^2)\ntheta = Arctan ((y-b)/(x-a))\n\nwhere (x,y) is any point on the circle, (a,b) is the circle's center, and r is the radius.\n\nThe formula for theta that I just provided may be precisely what you are looking for.  It works simply by first finding the "slope" of the line connection (x,y) and (a,b), then taking the inverse tangent of that slope, which 'converts' it to an "angle."\n\nHope this helps!
970313
r=radius\nd=diameter\n\n1 C=2r(pi)\n2 C=d(pi)\n\n(Same thing because 2r=d)
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
735905
a). If it touches the x-axis, then y is zero.  \nSo x^2 + 0 + 0 = 7\n\nx^2 = 7\n\nx = sqrt 7 or - sqrt 7  \n\nThe points are (sqrt 7, 0) and (-sqrt 7, 0)\n\nb).  Differentiate implicitly like this:\n\nd/dx x^2 + d/dx xy + d/dx y^2 = d/dx 7\n\n2x + (x dy/dx + y) + 2y dy/dx = 0\n\nx dy/dx + 2y dy/dx = -2x - y\n\ndy/dx (x+2y) = -2x - y\n\ndy/dx = (-2x-y)/(x+2y)\n\nc,d).  Just plug in the coordinates from (1)
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.
822973
Since the foci are along the y-axis, the major axis of the ellipse will be along the y-axis.  Also, the center of the two foci is the center of the ellipse, which will be (0,0) or the origin.\n\nSo the only thing left to figure out is the length of the minor axis.\n\nThe major axis, minor axis, and focal length (distance from the foci to the center) are related by the Pythagorean theorem:\n\na² + b² = c²\n\nYou know that a = 26/2 = 13, c = 12, so b = 5.\n\nTherefore the equation is:\n\nx²/25 + y²/169 = 1
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!!!
981742
x² + 2x + y² - 6y - 6 = 0 --- Add 6 to both sides...\nx² + 2x + y² - 6y = 6 --- Add some parentheses...\n(x² + 2x) + (y² - 6y) = 6 --- Complete the Squares*...\n(x² + 2x + 1) + (y² - 6y + 9) = 6 + 1 + 9 --- Simplify the two expressions in the parentheses, and add the numbers to the right of the "=" sign...\n(x + 1)² + (y - 3)² = 16\n\nFormula: (x - h)² + (y - k)² = r² --- Where (h, k) is the center point of the circle, and r is the radius of the circle.\n\nUsing this formula, (-1, 3) would be the center, and 4 (= √16) is the radius.\n\n\n*Completing the square is an algebra technique, also used in many types of calculus. The essential objective is to reduce a quadratic polynomial in a variable in an equation or expression to a squared polynomial of linear order. This makes the equation easier to solve in many contexts. More information can be found here: http://en.wikipedia.org/wiki/Complete_the_square
838513
Sigh - first guy had it right, but didn't explain it all.  \nPi = 3.14- all you need for the problem\nRadius= half the diameter of your circle. You find the midpoint and measure from there to the side of the circle, in any direction. Most problems give you the radius. or at least show you the midpoint on the circle.\n\nSo, it is 3.14 x the amount of the radius x 2= the area of the circle.\n\nYou can do this in reverse as well, if you know the area but not the radius.
586962
(6y^2-3y)(y+7)-->(2y)(y-1)(y+7)\nif that helps
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!!
178452
y^2 + 3y + 6 -4y^2 - 6y simplifies to:\n\n-3y^2 - 3y + 6\n\nIt's a second-degree trinomial expression.