if the equation of a circle is (x-2)squared+(y-6)squared=4, does it pass through point (5,6)?
467836
Remember that (5,6) is a point in the xy-plane.  Graphically, the picture we have in our head is drawing a circle and seeing whether or not the circle goes through that point.  But if you're anything like me, you can't draw circles worth anything, so another way is to remember that the equation of the circle is just another way to describe all of those points.  That sounds kind of redundant, but keep it in mind.  What it means is that if I take ANY point on the circle, the x and y values of that point will fit into the equation of the circle and make it true.  (This is true not just for circles, but for any equations, be they for lines, parabolas, etcetera.)\n\nSo, the point (5,6) gives us our x and y values.  Plug them into the left-hand side of the equation, and we get \n\n(5-2)^2 + (6-6)^2 = 9, which doesn't equal 4, so the point is NOT on the circle.  Actually, this question would have been fairly easy to see from drawing a picture, too, but that's harder to do here!\n\nHope this helps.
738725
Are you sure you have the first formula right?  cos(0) is 1, so it reduces to r = 3 which is a circle of radius = 3\n\nFor the second equation, do you mean X^2 + Y^2 = 4Y?\n\nJust replace Y with rsin(theta) and X with rcos(theta) and solve for r....\n\nThe original formula cannot be right...\n\nIt reduces down to r= 4rsin(x) which is an invalid formula.\nAre you sure there is Y after 4?
367523
x^2 < 4, swap the < for a = for a sec.  That gives you x^2 = 4, and x = +/- 2.  So on a number line, draw an open circle at -2 and 2 (open circle because the original problem was < and not <=).  Now comes the logical part - if you plug in something to the far left of -2, does it satisfy your original equation?  Nope.  How about far right of 2?  Nope again.  What about between -2 and 2?  Yep.  So fill in or darken the line between the open circles.
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.
1167592
6√2\n\nRemember it's a square so the sides are all the same length, and the area is s^2\n\nso s^2=36==>s^2-36=(s-6)(s+6)=0 so s is either 6 or -6.  -6 makes no sense here, so it is obviously 6.\n\nNow, since we have a square we can "draw" (imagine) a line at the diagonal, and that makes a right triangle (since the corners of a square are all 90º) with legs 6 and 6.  using the Pythagorean relation, we have d^2=s^2+s^2=6^2+6^2=72.  Thus d^2-72=(d-6√2)(d+6√2)=0.  Thus d is either 6√2 or -6√2, and again is obviously the positive one :)
324456
give parameters for x, y, and z.  Otherwise you cannot prove it, as no parameters are given for the equation as you described it.
737079
If the x2 mean x squared?\n\n4x^2 + 3x + 6
857047
don't know if you've taken Calc I or not, but if you have, differentiate your equation.  If not, you can look at the next paragraph.You'll get y'=-2x+6.  Remember that deriviatives tell the slope of a function at the x input.  Also remember that quadriatics are U shaped with the co-efficient of the squared term telling if the equation opens up or down.  (positive up, negative down)  Since it's negative, the equation will open down and have a maximum point. (the slope of the max point will be 0) so y'= -2x+6=0 (x=3) will give max value (3,9)\n\nSet your equation equal to zero and solve.\n(-x^2+6x=0)\n-x(x-6)=0\nx=0,6\nsince quadriatics are symetrical around the max/min point, find the point half way between 0 and 6 (x=3) and this will yeild your max/min point. (3,9)
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)
97164
5 x squared would just be 5x^2\n\nx^-2  =  1/x^2\n\nSo, y negative 6 exponent  =  1/y^6
434174
If the equation is y = 4x^2 + 4x - 5, then:\n\nRewrite the equation and complete the sqaure on the right side of the equation...\n\ny = 4(x^2 + x) - 5\n\ny = 4(x^2 + x + 1/4) - 5 -4(1/4)\n\ny = 4(x + 1/2)^2 - 5 - 1\n\ny = 4(x + 1/2)^2 - 6\n\ny - (-6) = 4(x + 1/2)^2\n\ny - (-6) = 4[x - (-1/2)]^2\n\nVertex for standard equation is in the form:\n\ny - k = +/- a(x - h)^2\nwhere:\nh = x-coordinate of the vertex.\nk = y-coordinate of the vertex.\n+ a = the parabola opens upward.\n- a = the parabola opens downward\na = the parabola will have a wide opening if a > 1.\na = the parabola will have a narrow opening if a < 1.\n\nTherefore vertex is at (-1/2, -6) and a wider opening which is upward.
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/
33948
u know general equation of circle is\n(x-h)^2+(y-k)^2=r^2\nwhere h and k are x and y coordinates of center of circle and r is radius of circle.\njust put the known values of h,k and r to get the equation.\nhey,i did it.
750847
you do not even need to complete the square,\nyour equation is the one of an ellipse\n\n9x^2-36x+4y^2=0\n9(x^2-4x+4)+4y^2=0\n9(x-2)^2+4y^2=36\ndividing by 36\n(x-2)^2/4 + (y^2)/9 =  1\n:D
965380
(x - 3)² + (y + 2)² = 4²\n(C): x² + y² - 6x + 4y - 3 = 0
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
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