Given a circle on the coordinate plane, Sal finds its center and its radius.
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- instead of eye balling the center, how do you find the center if you only given a point on the circle. Trial and error?(32 votes)
- If you have one point, and you do not know anything else, there are an infinite amount of circle centers that could be scattered about the Cartesian plane. Even with two points, there are still an infinite amount, though in a line equidistant between the points now. You need to know three points on a circle to define it. We could also know the center point and another point on the circle, which is what we are doing here. Sal 'eye-balled' the center and was given a point on the circle, which he then used to find the radius. If we do not know the center and are given a point on the circle, we are no better off than before.(47 votes)
- conic sections aren't functions, right? They have more than one corresponding point for x. Yet a parabola is indeed a function... (?)(12 votes)
- You're right: Conic sections are a bit like trees or rocks: There are all kinds of them, and they don't all look alike, or act alike. Whether or not a particular curve (conic section or not) is a function will depend on how we write its equation, or how we situate it on a coordinate plane. The equation of a closed curve (like a circle or an ellipse) can never be a function (at least: I don't think so. Can you say why?). The equation of any parabola or hyperbola can be written as a function, but there are equations of parabolas and hyperbolas that are not functions. For example, if we think of y as a function of x, then x^2 = y is a function, but x = y^2 is not, even though the two equations have congruent graphs.(11 votes)
- how we find the center of a circle if it cannot be eye balled?(4 votes)
- If you do not have the formula of it, you need at least three points you can eyeball on the circle, not inside. If you cannot tell any points exactly then there's no way to get the center.
With the three points you have, there are a few ways to do it, this is the way i prefer. You need to find two perpendicular bisectors (I can explain this if you need.) It doesn't matter which points you do, but you need to find two different perpendicular bisectors. Then, wherever these two intersect is the center of the circle.
Let me know if that doesn't help, of again if you would like an explanation on perpendicular bisectors.(5 votes)
- Instead of using the distance formula couldn´t you just count up from a point on the same line on the center?(1 vote)
- In some problems you could, but often the circle does not intersect at an exact point, so you would have to guess where it intersects. The distance formula generally gives you a more accurate answer. For example, in this video you would be able to count that the radius is about 2.25 units, but you would not be able to count that it is exactly √5 units.(6 votes)
- how to determine the radius if the circle has its center?(1 vote)
- What's the distance formula again? Was it sqrt(x_2-x_1)^2+(y_2-y_1)^2(1 vote)
- Was the radius that we solved for, an exact value or just an approximated?
cuz the result turns out to be only the "sqrt" of some number.. not any other number(0 votes)
- The radius, which turned out to be √5, is an exact number when written that way; it is not an approximation. If it had been written as a decimal, it would have had to be approximated because irrational numbers in decimal form never end and never repeat.(4 votes)
- how to determine the radius?(0 votes)
- So if you know the radius, then you know that pi*(radius^2) equals the area, so if we know the area, then we could just do a little bit of rearranging of the equation, here's the steps I went through to get to an equation which if you plug in the area of a circle, you can find the radius:
Area = Pi * radius^2 || divide pi by both sides
Area/Pi = radius^2 || take the square root of both sides
sqrt(Area/Pi) = radius || and there you go, we got that the radius is equal to sqrt(Area/Pi).
I haven't tried that out myself ever, but it should be right, hope this helps,
- ncochran2(2 votes)
- 1:401) I dont get why the change in x is 1 unit to the right, why couldnt it be one unit up.
2) why is it called CHANGE in x/y nothing is changing.
3) distance is just the radius right?
- The change in x has to be to the right or left, because the x-axis is horizontal.
The "change" in y or x just means how many spaces apart the point and the center are (the number has changed).
Yes, the distance is same as the radius.(2 votes)
- I thought the distance formula is the square root of (x2-x1)^2 +(y2-y1)^2(0 votes)
- Sal's version is the same. He just hasn't applied the square root yet. Notice, he has "d^2", not "d".
Note: The triangle in front of X and Y means "delta" X and "delta" Y. This is the same as saying the change in X = (x2-x1) and the change in Y = (y2-y1).
Hope this helps.(1 vote)
- [Voiceover] So we have a circle right over here and the first question we'll ask ourselves is what are the coordinates of the center of that circle? Well, we can eyeball that, we can see, look, it looks like the circle is centered on that point right over there and the coordinates of that point, the x-coordinate is negative four and the y-coordinate is negative seven. So the center of that circle would be the point negative four comma negative seven. Now let's say, on top of that, someone were to tell us, someone were to tell us, that this point, negative five comma negative nine is also on the circle. So negative five comma negative nine is on the circle. So based on this information, the coordinate of the center and a point that sits on the circle, can we figure out the radius? Well the radius is just the distance between the center of the circle and any point on the circle. In fact, one of the most typical definitions of a circle is all of the points that are the same distance, or that are the radius away from another point, and that other point would be the center of the circle. So how do we find out the distance between these two points, between these two points? So the length of that orange line. Well we can use the distance formula, which is essentially the Pythagorean theorem. The distance squared, so if the length of that is the distance, so we could say the distance squared is going to be equal to, is going to be equal to our change in x squared, so that right there is our change in x, and I have to write really small, but that's our change in x, our change in x squared, plus our change in y squared. Our change in y squared. Change in y squared. Now what is our change in x? Our change in x, and you can even eyeball it here, it looks like it's one, but let's verify it. It doesn't matter which one, just use the start or the end, as long as you're consistent. So, let's see, if we view this as the end, we'd say negative five it'd be negative five, minus negative four minus negative four And so this would be equal to negative one. So when you go from the center to this outer point, negative five comma negative nine, you go one back in the x direction. Now the actual distance would just be the absolute value of that, but it doesn't matter that this is a negative because we're about to square it and so that negative sign will go away. Now what is our change in y? Our change in y. Well, this is the finishing y, negative nine minus negative seven, minus our initial y is equal to negative two. And notice, just to go from that point, that y to that y, we go to negative two. So actually, we could call the length of that side as the absolute value of our change in y. And we could view this as the absolute value of our change in x. And it doesn't really matter because once we square them the negatives go away. So our distance squared, our distance squared, I really could call this the radius squared, is going to be equal to our change in x squared. Well, it's negative one squared, which is just going to be one plus our change in y squared, negative two squared is just positive four. One plus four, and so you have your distance squared is equal to five or that the distance is equal to the square root of five. And I could have just called this variable the radius, so we could say the radius is equal to the square root of five. And we're done.