How to graph a circle given the center and either the radius or another point on the circle.
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- what should be the exact angle between the plane and cone to get a parabola?(17 votes)
- Basically if you have a cone like and you slice it straight in half parallel to the ground you get a circle. Like slicing a cone on the line "A". But if you slice it perpendicular to the ground you get a hyperbola. If you slice it on a slan (0 to 90 degrees) t you can either get a ellipse or a parabola. An ellips will be from cutting the cone near the top so you can get the full circle. If you cut it on a slant near the bottom section it will be like getting an unfinished ellipse or parabola.(7 votes)
- Can a circle be considered a degenerate form of an ellipse?(5 votes)
- Yes. To quote Wolfram Alpha, the circle is a degenerate form of an ellipse as the eccentricity approaches 0.(6 votes)
- when I try solving the problem, I can not extend the circle I order to make the radius larger(4 votes)
- Make sure the inner circle is not "highlighted" or in its larger size. Then grab the outer circle with the cursor and see if it will extend. It took me a while to get the hang of it, but it does work. Good luck!(4 votes)
- how do I add another point on to the circle while graphing(3 votes)
- my video says "you earned 745 energy points" "replay video" but it also shows that I have only watched 3/4 of the video.(3 votes)
- Sounds like you need technical support. You need to go to the Help Center (a link is at the bottom of any KhanAcademy screen.(4 votes)
- Is this on a cartesian plane? why is it that I can't graph ellipses on my graphing calculator? Is it because they aren't functions?(3 votes)
- This is graphed on the cartesian plane.
Circles and ellipses are not functions, so graphing calculators will not usually be able to graph the equations directly. However, you can split the circle/ellipse equation into the top and bottom halves which are functions and can be graphed.(3 votes)
- find the eq of the tangent drawn from the origin to the circle (x*x)+(y*y)+(2gx)+2fy+c=0 and hence deduce a condition for these tangents to be perpendicular(3 votes)
- the point (0,4) for the circle radius was given but why wasn't the circle on the x axis point zero? sorry, just a bit confused(2 votes)
- The center is just a point, the coordinates for a point is simply (x,y), how much you move on the horizontal and how much you move on the vertical axis, so (0,4), you move 0 units on the x axis and 4 units on the y axis, that's why it is a up there.(3 votes)
- How can I construct a table of values?
Am I the one to contruct it?(2 votes)
- A table of values for graphing a function? It usually consists of two columns. One may be headed "x", and the other f(x). Under x, you list different values for x. Then plug each of these values into the function and the output value will be f(x). Make sure you keep each x lined up next to its corresponding f(x) value.(2 votes)
- At1:44, he could have used the distance formula to find the radius, and the equation of the circle to help. Why didn't he us it?(2 votes)
- Simply because the question didn't require him to go and make an equation, and it's faster without it. Using the distance formula to create an equation and graphing based on that is a valid solution, but lots more tedious than just plotting the circle's center and using a point the question tells you is on the circle to draw the circle.(2 votes)
- [Voiceover] We're asked to graph the circle which is centered at 3, -2 and has a radius of 5 units. I got this exercise off of the Khan Academy graph a circle according to its features exercise. It's a pretty neat little widget here, because what I can do is I can take this dot and I can move it around to redefine the center of the circle so it's centered at 3, -2. So, X is 3 and Y is -2, so that's its center. It has to have a radius of 5. The way it's drawn right now it has a radius of 1. The distance between the center and the actual circle, the points that define the circle, right now it's 1. I need to make this radius equal to 5. If I take that, so now the radius is equal to 2, 3, 4 and 5. There you go. Centered at 3, -2, radius of 5. Notice you go from the center to the actual circle it's 5 no matter where you go. Let's do one more of these. Graph the circle which is centered at -4, 1 and which has the point 0, 4 on it. Once again, let's drag the center, so it's going to be -4, X is -4. Y is 1, so that's the center. It has the point 0, 4 on it. X is 0, Y is 4. I have to drag, I have to increase the radius of the circle. Let's see, whoops, nope, I want to make sure I don't change the center. I want to increase the radius of the circle until it includes this point right over here, 0,4. I'm not there quite yet, there you go. I'm now including the point 0, 4. If we're curious what the radius is, we could just go along the X axis. X = -4 is the X coordinate for the center and we see that this point ... This is -4,1 and we see that 1,1 is actually on the circle. The distance here is you go 4 then another 1, it's 5. This has a radius of 5. But either way, we did what they asked us to do.