This task is usually a chore -- not anymore!
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For 30 years, I’ve waited for this video. The FAQ for the old newsgroup, comp.graphics.algorithms has this solution:
Subject 1.04: How do I generate a circle through three points?
Let the three given points be a, b, c. Use _0 and _1 to represent
x and y coordinates. The coordinates of the center p=(p_0,p_1) of
the circle determined by a, b, and c are:
A = b_0 - a_0;
B = b_1 - a_1;
C = c_0 - a_0;
D = c_1 - a_1;
E = A*(a_0 + b_0) + B*(a_1 + b_1);
F = C*(a_0 + c_0) + D*(a_1 + c_1);
G = 2.0*(A*(c_1 - b_1)-B*(c_0 - b_0));
p_0 = (D*E - B*F) / G;
p_1 = (A*F - C*E) / G;
If G is zero then the three points are collinear and no finite-radius
circle through them exists. Otherwise, the radius of the circle is:
r^2 = (a_0 - p_0)^2 + (a_1 - p_1)^2
Reference:
[O' Rourke (C)] p. 201. Simplified by Jim Ward.
This was my first exposure to analytic geometry when I was 12 years old, and led me to study with the University of Oslo via newsgroups before many Internet services had spun up. Great pick! Here’s the link:
www.faqs.org/faqs/graphics/algorithms-faq/
You can also get the equation for the circle with diameter AB as follows. Recall that the angle in a semicircle is 90, so the vectors (x-a, y-b) and (x-c, y-d) are perpendicular, whence their dot product is 0, which gives us the equation.
Indeed you can, and Penn uses this shortcut throughout his examples. The main task isn’t to find a circle with diameter AB, though, but to find the circle through the three given points.
I'm confused. At 4:30 he sets alpha = 2 and gets the equation (x-1)^2 + y^2 = 5, but shouldn't it be x^2 + (y-1)^2 = 5? That shifts the circle to the right one unit instead up one unit from the original. Same for when alpha = -2 at 5:00 minute mark.
I saw that too. The graphs of the circles don't match the equations.
yeah, i saw it too. i has to be x^2 + (y-1)^2 = 5
sign. it is very obvious mistake that (x-something) is shifting left and right, not up and down.
You're right: he switched x and y
@@ethancheung1676It’s very obvious, but not really impactful to my overall understanding once it’s pointed out
Another approach is with matrix determinants. Define a 4x4 matrix that's a function of indeterminates x and y. The first row is [x^2 + y^2, x, y, 1] and the the next three rows are the same as the first except you plug in your three points for x and y.
Then set the determinant equal to zero to get an implied equation in x and y. The equation is obviously satisfied when x and y are one of the three points since that makes two of the rows identical.
Moreover if you expand the matrix determinant along the first row you see this has to be the equation of a circle (or degenerate version thereof, like a point).
So just take the appropriate subdeterminants to extract the coefficients that make up the equation for a circle.
AB is not restricted to be a diameter of the final circle, but sets the restriction on the family of circles having chord AB.
That is a neat way to summarize what is happening
Thank you. That wasn't clicking for me.
He really should have made that clear at the start
Take two pairs of points, construct perpendicular lines on each center so they intersect. Works fine geometrically and mathematically.
Easy to understand as well.
Y=(A+B)*0.5 + p+t{A-B)
X=(C+B)*0.5 + q+t(C-B)
with vector t((a, b)) = (-b, a) {rotated 90*)
This can be solved for p or q to get the center. The distance from any point to the center is the radius.
You get some p=n/d. If d is 0 there is no solution (points on a line).
Interesting that this also shows |ℝ| = |ℝ x ℝ| since you have a mapping from points on the plane to a single real number α. (I don't think this is a common way to prove that lol)
Nah the map is not a bijection. Each alpha does not give you a particular C but almost a whole circle of possible C's.
4:30 It is y-1 and not x-1 (also y+1 and not x+1)
11:44 You forgot the Y0 in the denominator of the last term of the LHS and Y0^2 accordingly on the RHS
Thank you! I was looking at those equations and I could not for the life of me figure out how x^2 + y^2 -4 = 2y could be transformed into (x-1)^2 + y^2 = 5. I figured it had to be a mistake of switching the variables but was too lazy to do the calculation myself.
Interesting, my first thought was to just find the bisector of 2 segments and where they intersect (circumcenter and therefore center of inscribed circle), calc distance between one of the points and the circumcenter (radius of circle) and get the equation via (x-h)²+(y-h)²=r²
I find this highly interesting, why at school certain algorithms to calculate something are preferred to other methods. I never heard of this method before and do love its elegance. Yet I can see why one puts students through the chore of solving the system of equations you get by simply pluggin in the points into the circle equation, if your ultimate goal is to give them a secure way to calculate the circle equation with familiar tools.
Here's the point what we certainly do wrong when teaching math: The goals to be fulfilled are too algo-centric. Students are way too often formed into little iterative computer programs, that can fufill certain calculations by following algos without widening their mind really.
I mean, who cares, if anyone can reliably calculate circle equations out of three points. We rather should try to teach them the beauty of approaches like this one, that seem unfamiliar and maybe don't stick, but if you are just for a moment able to appreciate the beauty in it, that's way more valuable.
I was literally thinking about the process of fixing a circle from 3 points for a research problem today: this is so well-timed!
My only problem with this is that I don't expect to ever want to find the equation of a circle given 3 points. Bugger!
I feel like this is conceptually and physically more complicated than just finding the intersection of two of the perpendicular bisectors of the triangle...
This would be an interesting method of inscribing Bresenham's circle algorithm in a rectangle. The current implementation uses a center point and radius, which when using integer arithmetic, causes problems with centering the circle on a pixel grid in so that it correctly inscribes the rectangle. By using this method, you could set A and B to be the left and right edges of the rectangle at any y value within the bounds of (rectangle top -> bottom) and C to be the top of the rectangle with any x value within the bounds of (rectangle left -> right). 🤔 Hmmm.... Giving this some thought, A, B and C wouldn't even need to be points, just the x extents and one y extent.
I'm going to have to play around with this.
This can be argued in more abstract terms without the need for all the algebra.
The diameter g(x,y)=0 is just the limiting case of a second circle through A and B, so imagine you have two circles instead of a circle and a line.
It can be shown that the set of equations of all circles through A and B satisfy all the requirements for a vector space, most notably closure under addition and scalar multiplication. It can also be shown that the equations of two distinct circles are independent and span the space, so they are a basis for the space.
The equation a.f(x,y) + b.g(x,y) = 0 is just a linear combination of the basis vectors, so gives every vector in the space, that is every circle through A and B.
Dividing by _a_ and rearranging gives f(x,y) = c.g(x,y). Now we have eliminated all linear combinations where the first circle is counted zero times, so we can no longer get the second circle. Or in terms of the original problem, we cannot get the equation of the chord AB, which is what prevents the three points from being collinear.
3 isoceles triangles with the center point created by intersecting the undefined points.
Wow - circles are complicated in an x ,y, z world (and let's not get started on ellipses). It's like the universe doesn't care about our "3D" view? :) Lovely vid MP - very satisfying.
Construct two perpendicular bisectors between two sets of the three points. Where they intersect is the center of the circle.
Or if you construct THREE perpendicular bisectors and the drawing isn't 100% accurate (likely with real exercises) you get a little triangle. Put the point of your compass in the middle of it and it minimises the error.
@@zh84 in the context of the video you could calculate equations for the bisectors which is easy enough and solving a system of 2 linear equations is even easier. Bam, the center.
great video! THank you!
very nice method. I'm not sure if this is connected but I remember there was another problem. Given points (x1,y1), (x2,y2) ... (xn, yn) find the circle containing all of them (points can lie inside) with smallest radius. For me it was unsolvable at a time with algorithm that has decent asymptotic complexity.
Alternatively, we can have the equation from the circle inscribed chord angle theorem and cosine formula : circle is locus of points P such as squared cosine of angle APB equals squared cosine of angle ACB. Moreover comparing signs of cosines angles APB and ACB we know which side of the line AB, P belongs to ;-) I know a similar method for conic when you know equations of two tangents l1(x,y) = l2(x,y) = 0 and a point (x0,y0) on the conic, then equation of conic is simply l1(x,y)*l2(x,y) + alpha = 0 where alpha is determined by substitution (x,y) = (x0,y0) ...it is a method using "pencil of conics".
I think you might be misremembering that last part: two lines and a point are not enough to determine a conic uniquely. l1*l2+α=0 defines a family of hyperbolas with common asymptotes l1 and l2, not a general conic. The asymptotes are tangent to the hyperbolas at infinity, so I’ll give you that.
As an interesting complication if you tried to train a symbol manipulator to do this, if F(x,y)=0 is the equation of a circle or line or some other shape, generally, F(x,y)^2 = 0 is also the equation of that same circle, line or whatever other shape. So you'll run into trouble communicating the F(x,y) = alpha G(x,y) step if both functions can just be squared.
For A=(a, b) and B=(c, d)
you get
f(x, y)=(x-a)(x-c)+(y-b)(y-c)
analogously, for g(x,y) you can use
g(x,y)=(x-a)(y-d)-(x-c)(y-b)
h_1(x,y)=(x-a)(x-c)-(y-b)(y-d)
and h_2(x,y)=(x-a)(y-d)+(x-c)(y-b)
are different hyperbola.
You should be careful, though, in speaking of "the" equation of circle and line. This only works if f(x, y) and g(x, y) have a specific form: for instance, f should be quadratic and g should be linear in its arguments.
Otherwise, a maverick might define f(x, y) = (1 + x^2) (x^2 + y^2 - 1) and g(x, y) = exp(x) (y + sin y); the sets { f(x, y) = 0 } and { g(x, y) = 0 } are still your circle and line, but the equations f(x, y) = α g(x, y) result in rather peculiar curves...
I prefer to find circle equations using linear regression. It's really easy to derive and to apply, can find a reasonably well fitting circle through more than 3 points that do not exactly lie on a circle, and generalizes straightforwardly to spheres of arbitrary dimension (and less straightforwardly, arbitrary codimension).
This was a chore indeed.
Thank you, professor bb
Let A(x_a,y_a), B, C be the given points. Let E(x_e, y_e) be the midpoint of AB and F(x_f,y_f) be the midpoint of BC.
Let m_e be the slope of line AB and let m_f be the slope of line BC. Then, line e, the perpendicular bisector of AB, has equation y-y_e = -(x-x_e)/m_e, and line f, the perpendicular bisector of BC, has equation y - y_f = -(x-x_f)/m_f.
Solve the system of equations of lines e and f. This is the center of the circle C(x_c,y_c). Then, r^2 = (x_c - x_a)^2 + (y_c - y_a)^2.
This reminds me of a Lagrangian.
0:10 three different and nonlinear. Well unless we consider a straight line a circle with radius going off to infinity
The equations written after 4:30 are of circles which move left and right, not up and down. It just isn't a good demonstration of the method if the algebra is so wrong.
Another cool result .
Can we do the same thing but on parabola, hyperbola , ellipse?
Why does this remind me of the Lagrange multipliers method? Is there a connection?
You could consider this also falling into the calculus of variations. Or alternatively you can rephrase the "find a circle through three points" problem as an optimization problem.
While these videos are often fun and clever, I really find the frequency of really glaring errors pretty disappointing. Some older videos at least used to have edits with disclaimers (oops, that x-1 should have been y-1) but not recently
Hm. Not bad. But I think, that my approach is a tiny bit more elegant than that approach.
Obviously and trivially the equation for the *perpendicular bisector PB for two given points A and B* is given by the following expression and equation:
For any X∈ℝ² and X∈ *PB(A,B): (X-(A+B)/2)·(B-A)=0* with "·" being the scalar product between the two vectors here between X-(A+B)/2 and B-A.
It should be clear, that for any such point X∈PB(A,B) the distance between X and A d(X,A) should be the same as the distance between X and B d(X,B), such that *d(X,A)=d(X,B) for any X∈PB(A,B).*
*So further more for any given three distinct points A, B and C (not all three simultaniously being on the same line) any and every perpendicular bisectors PBs between two of these three given points A, B and C will be distinct AND NOT PARALLEL to each other, such that they cross and intersect each other in a specific, particular and unique point M, such that d(M,A)=d(M,B)=d(M,C)=r.*
Using your example: A(5|3); B(-2|0); C(1|1)
PB(A,B): (X-(A+B)/2)·(B-A)=0 ⇔ 2(B-A)·X=B²-A²
⇒ (I): -14x-6y=-30
PB(A,C): (X-(A+C)/2)·(C-A)=0 ⇔ 2(C-A)·X=C²-A²
⇒ (II): -8x-4y=-32
⇒ 3(II)/4-(I)/2: x=-9; in -(I)/4: 2·(-9)+y=8 ⇒ y=26
⇒ The middle point of the circle with the three given points A, B and C is M(-9|26).
The radius r is then given by r=d(M,A)=d(M,B)=d(M,C)
=((M-A)²)^0.5=((M-B)²)^0.5=((M-C)²)^0.5
=(14²+23²)^0.5=(7²+26²)^0.5=(10²+25²)^0.5=(725)^0.5
And with those we obviously get the following equation for the circle with the three given points A, B and C:
Cir(A,B,C): d(X,M)²=(X-M)²=r²
Here Cir(A(5|3),B(-2|0),C(1|1)): (x+9)²+(y-26)²=725.
Done!
Edit: Yes, you can generalize this to higher dimensions:
For any given n+1 distinct points (not all n+1 points simultaniously being on the same hyperplane) any and every perpendicular bisectors PBs between two of these n+1 given points (, which are in general hyperplanes of ℝ^n by their definition of *PB(P,Q): (X-(P+Q)/2)·(Q-P)=0* ) will be distinct AND NOT PARALLEL to each other, such that they cross and intersect each other in a specific, particular and unique point M, such that d(M,P1)=d(M,P2)=...=d(M,P(n+1))=r.
I was wondering if he was going to limit the points to being non-collinear or if the line was going to be a the limit of a circle as the radius approaches infinity.
Love when you teach geometry
Are you kidding me. I literally had to do this at work THIS WEEK! I used triangle circumcircle.
My version was easier 😂
When building the result by incorporating point C, will there be a tell while working to tip off accidentally choosing a C that actually is co-linear to A and B?
All points on the line satisfy g(x,y)=0, so if C is on the same line then f(x,y)=αg(x,y) reduces to f(x,y)=0. The tell is the disappearing RHS.
I loved this video! 😊
y nought? because yes
Judging from earlier posts it seems that there are a few ways to solve this using different techniques
Now one of potential restrictions on a formal Math course is students are assessed on learning outcomes and that will usually mean using the "taught" methodology.
Wouldn't it be wonderful for a happy postgrad or anyone(?) to bundle these solution methods together?
For why?
Potential answer: sometimes there are too many restrictions in taught math that obscures big picture of multiple ways of doing things in math?
Besides. once transformed will points A'' and B'' be equidistant from origin noting, of course, that y=0 for both A'' and B''?
why don't u bundle these solution methods together?
@@charleyhoward4594 I sometimes do for personal use and I do those in a way that is not really intended for public broadcast>
Making it a RUclips event requires skills not only in math but also in presenting math and doing such things digitally.
An interesting challenge all the same
there are a great many things awry with this video
Nice trick.
I like the method, but unfortunately it is only "quick" for the special cases when the two points are lying on one of the axes. With all the other transformations and back substitutions, I do not consider it as a "quick" method.
The transformations were only to reduce all cases to 1 special case, to prove that the whole thing works for any 3 non-collinear points. As you can see when he works the example, you don't need them for the actual solution.
I can't believe I've never seen this method before. I've had problems where I've had to find the equation of a circle passing through 3 points and I wish I'd known this method. How many points do you need to fully define a sphere? Is it 5?
I think 4 is enough. If you start with 3 points in space, they determine a circle. Then if you take the line perpendicular to the circle at its center, that line will contain the center of the sphere. Doing that with another 3 of the 4 given points gives a second line, and the intersection of those lines will be the center of the sphere.
I suppose in general, it takes n+2 points to determine an n-dimensional sphere? It does work when n=0 as well, since a 0-dimensional sphere is a pair of points (defined by the equation (x-a)^2=r^2).
4
The number of independent variables in an equation determines the number of degrees of freedom in choosing values for them. So for a sphere, whose equation looks like x^2 + y^2 + z^2 = r^2, you have four independent variables, and hence 4 points will determine the equation (subject to constraints such as non-collinearity).
@@RexxSchneider Ah, I should have thought of that. If you have 4 equations of the form
(x_k-a)^2+(y_k-b)^2+(z_k-c)^2=r^2, then subtracting pairs of them gives 3 independent linear equations. Then that allows you to get everything in terms of just one variable, and plugging that back into one of the original equations solves the system.
JEE kids -laughing- crying in the corner
Why is g(x,y)=y the x-axis?
y=0 is the x-axis.
g(x,y)=0 is the equation of the line. well g() is the function and the line is where g()=0.
y=0 is the x-axis.
So, g(x,y)=y is the function where the x-axis exists everywhere that g(x,y)=0
Y=0
Okay, for y=0 this is the x-axis! I got it, thanks
𝑓(𝑥, 𝑦) = (𝑥 − 𝑎) (𝑥 − 𝑐) + (𝑦 − 𝑏) (𝑦 − 𝑑) is *a* function s.t. 𝑓(𝑥, 𝑦) = 0 gives *an* equation of the circle with a given diameter. But it is not *the* function with this property. 𝑓(𝑥, 𝑦) = exp((𝑥 − 𝑎) (𝑥 − 𝑐) + (𝑦 − 𝑏) (𝑦 − 𝑑)) − 1 works as well. The same goes for the equation of the line. Will the statement that 𝑓(𝑥, 𝑦) = 𝛼 𝑔(𝑥, 𝑦) gives the equation of a circle at all for some 𝛼 be true for any such functions?