Can a line intersect a circle in at most one point?
602663
No, a line cannot intersect a circle in at most one point, unless you're talking about tangent. A line can be tangent to a circle at only one point.
823012
Either one is possible. Consider y=x² (one critical value, which is not an inflection point) and y=x³ (one critical value, which IS an inflection point).
268937
No. There are 360 degrees in a circle.
1124334
two circles, in parallel planes with the line connecting their centers forming right-angles with each plane in all directions. ... and the walls comprising the set of line segments one of whose end points is on one circel and the other end point on the other circle withe the property that the segments at each point on circle(A) is the minimum distance to circle(B).\n\n.. or, since you use the word "cylinder" in the question:\n\na cylinder, each of whose end-faces is a circle.
1072716
If you mean the line is level to the surface of the Earth all the way around, then no, it's approximately a big circle.
508642
Well you can't really have a center of the diameter line of a sphere.  But I guess the common dividing point would be where the prime merdian (zero degrees longitude) passes through the equator.  The intersection point is in the eastern Atlantic ocean, south of Ghana in Africa.
1131299
Well, a circle does have a side.  It is a continuous set of points that are all equidistant from a common point, thus giving it one side.
68097
do you mean things as far as max and mins, asymptotes and things like that?\n\nif you are a little more specific i can help you better, or e-mail me, it would be a little easier  matttlocke@yahoo.com
64832
You need to know the distance between the centers, because the circles could have different placements\n1- one inside the other\n2- doesn't intersect\n3- partly intersect
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)
833623
perpendicular distance s the shortest\nlet the pont be (x,y)\ncentre of the circle=(1,2)\nequation of the line joining (1,2)and (-2,-2)\n(y-2)/(-2-2)=(x-1)/(-2-1)\n-3(y-2)=-4(x-1)\n4x-3y+2=0\nsolve this with the equation of the circle\nto get the coordinates 'x' and'y'
267152
any general line passing through the mid point of the given line segment.
225717
Yes. Think of the trignometric circle (radius = 1 and centered at the orgin (x=0, y=0). The so called tangent, is the line x=1 (that *is* tangent to this circle). The numeric value that we call tangent of an angle (alfa) is the y value of the intercept of a line passing thru the origin and forming an angle = alfa with the x-axis.\nIn other words, draw a trignometric circle and a verticle line tangent given by x=1. Draw another line forming an angle w/ the horizontal axis and passing thru the origin an stretch it so that it crosses that line. The height of the point of intercept of these two lines is the value of the tangent.\n \nThe co-tangent is the same but w/ y=1.\n \nEdit:\nThe secant is the distance from the origin to the intercept in the explanation above. Since it is a segment of the diameter, it is over a line that separates the trig circle, and is thus a secant.\n\nCheck out this link (skip the beginning of the page):\nhttp://www.npcts.edu/math/PreCalculus/Transcendental/Trigonometric/Definition/index.html
401318
No, the only requirements for an inscribed circle are that the vertex has to be on the circle, and both of the rays have to intersect the circle.
149565
cool! I'll think about it\n\n---\n\nhere I am.\nUse the two perpenducular segments as a coordinate system with O on their intersection.\n\nSay: the 4-3 segment defines the x axis, and the 2-6 segment defines the y axis.\nYou know four points belonging to the circle: (-4;0),(0;-2),(3;0) and (0,6)\nThe center of the circle has the mean value between -4 and 3 as its x and the mean value between -2 and 6 as its y (property of chords: the center is on the chord's axis)\n\nso the center's coordinates are C(-1/2;2)\n\nNow you can easily calculate the distance between C and anyone of the four points you know to get the radius.\n\nThen finish finding the diameter\n\n----
151877
Theorytical if the line are formed on a flat gemetrical plane they will not intersect.Or if two tangent in a in the same circle are parallel they will never intersect.\nThe fith postulate of Euclid tackled this question.Later mathematicians, like gauss proved that they could intersect. But this proof was based on a different idea=paralell line having curvatures! Of course they would intersect. So the geometry which based on curvatures was started and was called Non Euclid Geometry.\nActually this could have been proven alot fast with some common sense! In reality it is imposible to construct a perfect circle or a perfect straight line(this is my own postulate).   This is because we are not able to have a perfect resolution of measurement.\nNot matter how you try to measure two parallel line  the results is ether convergence or divergence of the two lines if you make the line long enough!So in practice the 5th postulate of Euclid did have a practical application.
48873
well i would like to pretend its a regular hexagon, in this case you would have to draw a line from one vertex to an opposite one, the line should meet at the center of gravity (the point where all 3 lines intersect)