4 Squares 1 Circle

[@andonel]

i completely forgot about coordinate circle solving! very clever!

[@brianglendenning1632]

Me too!

[@jobaecker9752]

As did I. I immediately went into forming triangles. But lo and behold, coordinates to the rescue! How. Exciting.

[@rudychan8792]

Me too!^

[@gachanimestudios8348]

I legit had that as my first thought, lol.

[@transfer1992]

Same here, great to refresh the mind.

[@pixl_xip]

The double box was very exciting :)

[@FlyGuy2000]

The rare and elusive double box appears.

[@somebody-anonymous]

Underful hbox badness 10000

[@kaitek666]

this was one of the best ones so far. I absolutely loved the idea of putting it on an axis - first time seeing a solution like this. how exciting

[@starboy001]

The best and unique thing about Andy is that he shows each step how it is done so everyone can understand how it's done. Just like here he even give example of coordinate geometry of problem before using it. Thanks Andy

[@tsjbb]

how exciting

[@raysye4775]

hpw exciting

[@goodshiro10]

how exciting

[@zenir2591]

how exciting

[@giammaria_ezel]

how exciting

[@tusharbharti1445]

how exciting

[@SHAP00PIE]

Your channel made me like math and look at it differently. More like fun puzzles.

[@ldrgrm-venom_grim-pg3dandm64]

same

[@brycehins206]

I often describe my double major in phys/math as a 5 year long sudoku puzzle

[@osasunaitor]

Well done, that's the way. You can only tolerate maths if you see them as a series of clever puzzles to solve

[@rupom_1670]

Using coordinated plane here is really clever

[@jfm3258]

These videos make me feel speechless, he is not showing how to use numbers and formulas, he is teaching to think and to solve problems. HOW EXCITING!

[@quigonkenny]

It's even better when you try to come up with different ways to get the result. I used the law of cosines and the center angle/circumference angle theorem, as well as some distance formula.

[@jeffreypauk679]

I took the surveyor approach (being a surveyor and surveying instructor). Essentially took the bottom side of the bottom left square, extended it to right to intersect right side of rightmost square extended down. Established perpendicular bisector to left of rightmost square. From this, established right trapezoid with bottom base length of 16, right side perpendicular to base with length of 6, side from right side bisector left to radius point with a length sqrt(R^2-4), and a diagonal left side with distance of R. From radius point, construct line perpendicular to bottom base. This leads to a right triangle with hypotenuse of R, one side of 6, and another side of 16-sqrt(R^2-4). Plug into Pythagorean Theorem and end up with R^2=85. Area=85π.

[@way2patience670]

very interesting approach

[@jiminboo]

I love how the operations are in a different color. Visually makes things easier to follow, thanks!

[@akaRicoSanchez]

If you start with a little geometry and draw the perpendicular bisector of two of the circle's chords, you find the center of the circle very easily and from there a little use of Pythagorean theorem gives you the radius squared. Multiply by Pi and voilà :)

[@maxwellsdemon10]

Did you actually work it out like that? I totally concede that I might miss an obvious solution using only the Pythagorean theorem here, but I think you need more tools. I also thought it should be quite easily solvable and wanted to comment the easier solution to the problem here, but in the process of working it out I found out I needed some more tools and I have an argument why Pythagoras isn't enough. Assume P1,P2 and P3 are given as coordinates and lie on a circle. We can now do the perpendicular biscetion you talked about of the lines P1 P2 and P2 P3. Let's call the point the perpendicular lines cross the chord of the Circles S1 and S2. Now we have right triangles, where the radius is one length, the distance from the middle point to S1 and S2 and the Distance between S1 or S2 to P1, P2 or P3. The system of equations we get from Pythagoras now is not uniquely solvable. It cannot be, since we have ONLY used the distances between P1,P2 and P2,P3. These distances are not enough to uniquely define a circle. We need the information about the angle. If I use the law of cosines and compute the angle where the lines meet at P2, we can actually work it out, since we can quite easily get the angles in the right triangles between S1 P2 and the middle point. If we have these angles, we can get the radius, since we know two angles and the Distance between P2 and S1. But honestly the calculation is a pain and definitely harder than what was shown in the video. If I'm wrong, I would be super interested in a solution that works the way you have described it, but I found it to be way more tricky than it appears at first glance.

[@akaRicoSanchez]

@@maxwellsdemon10 I think your issue is only one of method. You are generalising the problem to find a solution for all possible P1, P2 and P3 and then try to plug in the specific P1, P2, P3 of the problem to find the solution of the problem. So basically you are building a nuclear missile to kill a mosquito :D If you work step by step using the specifics of the problem, it is dramatically easier. Let's have the bottom left contact point with the circle be P1 et let's set it at (0;0), making P2 (16;4) and P3 (16;8) since squares of area 16 have a side of length 4 (and I am using horizontal and vertical for my x and y axis). From this you can calculate the middles points S1 (8;2) and S2 (16;6). Very obviously the perpendicular bisector of the chord P2-P3 will be horizontal since the chord is vertical, and we know it will go through S2 and the center of the circle (let's call it C )so we know both have the same Y coordinate : 6. The other perpendicular bisector is slightly more complicated as it is slanted, but we know it's slant will be the negative inverse of the one of the chord P1-P2 since it's perpendicular. The slant of of P1-P2 is 4/16 or 1/4, so the slant of or perpendicular bisector is -4. Things work out very nicely for us because the bisector goes through S1(8;2) and we want to go up to the center C (?;6) and that's just 4 above, so we just need to go 1 back on the x axis, so C is at (7;6). Note that these previous two verbose paragraphs with coordinates take 10 seconds on graph paper if you draw the problem. So now with C, S2 and P3 we can go pythagoring: C-S2 is length 9, S2-P3 length 2, 9 squared plus 2 squared is 85 witch is the radius squared since C-P3 is a radius of the circle. so 85 Pi is our answer for the area of the circle! How exciting I guess :D

[@ldrgrm-venom_grim-pg3dandm64]

well what if it’s not to scale

[@feyyaznegus3599]

Yes, a geometric solution is also possible. You know where the horizontal diameter line is. The center of the circle is somewhere on this line. You draw your radius from the center to the left bottom corner, and draw a triangle using this radius as your hypotenuse. Then, you use the Pythagorean theorem: 6^2 + (16 - sqrt(r^2-4))^2 = r^2. Solve for r^2.

[@maxwellsdemon10]

@@feyyaznegus3599 Thanks, this was living rent free in my head. It's a nice solution, congrats, I couldn't work it out. For all the people like me who are bad at geometry, this is the detailed solution. We can take the two points that intersect the circle on the right side and draw the perpendicular line through the midpoint (This line is vertical, but obviously this works in any orientation). We know that such a line passes through the center. Let's call this diameter line d, the center C and the midpoint M. We can now compute the distance between M and C using pythagoreans theorem. We draw a orthogonal triangle between M, C and one of the intersections. Since M is the midpoint, the distance to the intersection is 2. So the distance between M and C is sqrt(r^2 - 2^2). We now draw a line from the center C to the point on the bottom left and add a vertical line (perpendicular to d). Let's call the intersection between d and the vertical line A. A, C and the point in the bottom left now form a orthogonal triangle with r as the hypotenuse. The vertical line has the length 6, since it has the length of 1.5 blocks. Now we only need the distance between A and C. We know that A and M are separated by 4 blocks, so the distance is 16. Subtracting the distance of C and M yields: 16 - sqrt( r^2 - 2^2). Now we can insert this in pythagoreans theorem and get: r^2 = 6^2 + ( 16 - sqrt(r^2 - 2^2))^2 Solvong for r yields r=sqrt(85)

[@MrMike3137]

Mind Blown. It's been so long but you still make it look so easy. This is one of my favorites. Rock On

[@BobY52944]

I love the way that you work the equations on the screen. Very smooth and easy to follow. How exciting!

[@richardl6751]

Food is always better when spicy, at least a little.

[@ExzaktVid]

Food is always worse when spicy, no exceptions.

[@richardl6751]

@@ExzaktVid Unlikely you will get a heart.

[@qyxyp]

what about candy? Do we count that to food?

[@richardl6751]

@@qyxyp Sure. Why not?

[@qyxyp]

@@richardl6751 because it‘s not better when it‘s spicy mostly

[@TheRaptorsClaw]

Finding k=2 could be done by deduction rather than mathematically, as a square with two points touching the interior of a circle will always have its symmetry line along the diameter. As the square is 4 units long, and we're defining 0,0 to be the bottom point, and 0,4 the top, the centre of the circle must thus be (h,2). Fun video, thanks for this one!

[@Orphanss]

really fun doing honestly

[@Geozeff]

Very cool solution! Also make the algebra a little less spicy by reading k=2 from the coordinate plane at 1:14 as that's the only way a square can fit like that

[@atrus3823]

I was thinking the same thing. And then you only need two points, which means, there is only one way a single square can fit like that. You can also see that we don't need 3 points, because once you know k = 2, (0, 0) and (0, 4) give the exact same information.

[@animextempleedits]

i dont understand can you explain why k=2?

[@Geozeff]

@@animextempleedits Since both vertices are touching the edge of the circle, the square must be as far right as it can fit, making it land nicely on the grid

[@paulvansommeren]

Because of the right square touching the circle twice, you can already deduce that k=2

[@trancozin130]

Can you explain?

[@paulvansommeren]

​@@trancozin130 You only need to consider the circle and the right square. Because it's touching the circle on both of the right corners, the rightmost extremity of the circle has to be right on the horizontal symmetry axis of the right square. If you imagine drawing a horizontal line through the middle of the square, you'll see that it goes through the right extremity of the circle. There's probably some proof for this, but to me it makes intuitive sense so it's hard to explain...

[@HR3EEE]

@@paulvansommerenwe’re gonna need some back up explaining here lol.

[@pikminman13]

@@HR3EEE the square is made of 4 straight, perpendicular, congruent lines. this means that the square is always "centered" if two of its (non opposite) corners are touching the edge of the circle. so you know the length of the square's side is 4, and thus half of that is 2. (were the opposite corners touching, you would instead have k be half the diagonal.)

[@FalconerCH]

@@pikminman13I dont understand, you mean after the circle is put on the plane?

[@londonalicante]

The rightmost square is centred symmetrically on the horizontal diameter of the circle (this has to be so since the chord it makes with the circle is vertical, but can be made clearer by drawing additional mirror flips of the other squares.) Given this symmetry, k is obviously 2 by inspection.

[@justforplaylists]

Clockwise from the top right, call the points ABC. The perpendicular bisector (p.b.) of AB is horizontal and halfway up the middle row. The perpendicular bisector of AC passes through the bottom right of the square in the second column, and goes left 1 units for every 2 units up. So they intersect in the middle of the middle row vertically, and one quarter of one square to the left of the middle vertical line. That's the circumcenter, and is 2 units down and 9 to the left of A on the circle. So pi r^2 = pi (2^2 + 8^2) = 85pi.

[@Pedrovosk13]

So satisfying to watch! The way you solved makes it look easy. Congrats and keep it up!!

[@jamesbod9539]

You jump right into it with no bs! I love it! Great vid and great channel

[@dozenal9420]

An alternative (and perhaps spicy!) solution: There's a lesser known theorem stating that abc=4RT, where a, b, c are the side lengths of a triangle, T its area, and R the radius of its circumcircle. The three points marked can form a triangle (where the circle in question is its circumcircle so this is the radius we're looking for), with side lengths we can work out to be 4, ~16,49, and ~17,89 using Pythagoras. As stated, the rightmost side has length 4 and the altitude to this side has length 16 (using that the squares have side length 4). The area T therefore equals 1/2*4*16=32. Putting all this together, R=4T/abc= sqrt(85). Finally, by our beloved πR^2, the area is 85π. No diagram needed!

[@HR3EEE]

Can you please explain the Pythagoras section of your proof.

[@roblowery3188]

Andy, It would be so cool to see how you develop a curriculum for all the "-not-so-often-used" formulae. I sit here befuddled by the problem and very often your method of attack seems new to me. Keep up the great work!

[@marvinochieng6295]

The channel i relax to everytime he uploads. Thank you andie

[@elmukaltino]

I came to the same answer just with the Circumcircle radius formula and while calculating the final result I found out that 85 is also 1+4+16+64, which is a lovely insight :)

[@creepx_hd1236]

This might be one of my favorite videos of yours. How Exciting!

[@valorandrews190]

If I had been taught math this way I would have retained and learned so much faster. These unique problems and explanations are perfect and make so much sense.

[@GhostStaff27]

This is incredible clever! I never imagined you could place the circle inside the Cartesian coordinates and then apply Analytical Geometry. Outstanding!

[@Dazzle_Novak_]

Right square touches the circle twice while being strictly parallel to X axis, therefore he's on the centre of X axis. If we mirror the down left square by the X axis we'll have ourselves an inscribed isosceles trapezoid with sides 4 and 12. Then we're just solving it by the formula.

[@therealawesomequest]

I feel betrayed, solving a system of equations is objectively the least fun way to solve any problem.

[@SassInYourClass]

This was really cool to do first and then watch you do it. I took a more geometric approach that gave me similar equations to solve. Ignoring all of the stuff I drew that was unhelpful, here’s what worked for me: All of the squares have the same area of 16. That means their side lengths are sqrt(16)=4. Three points define a unique circle. The center must be equidistant from all three points. That distance is the radius. I drew an approximate center with radii out to each point. Since the two points on the right are exactly vertical to each other, this center would be vertically equidistant between the two, meaning it was vertically 4/2=2 units higher than the lower right point. I also extended the “grid” formed by the squares, specifically horizontal line segments from the bottom-left square over to the right and across the bottoms of the two vertically middle squares. Then, I dropped a vertical line from the center down to the lowest horizontal line segment I drew. This formed two right-angle triangles, both with radii as the hypotenuses. The triangle on the left had a height of 6 and an unknown width of x. The triangle on the right had a height of 2 and an unknown width of 16-x. Using the Pythagorean Theorem, I was able to set r^2 equal to two different statements containing x. r^2=(6^2)+(x^2) r^2=(2^2)+((16-x)^2) I then set those equal to each other and solved for x. (6^2)+(x^2)=(2^2)+((16-x)^2) 36+(x^2)=4+256-32x+(x^2) -224=-32x x=7 I plugged x back into one of the equations to find r^2. r^2=(6^2)+(7^2) r^2=36+49 r^2=85 And then the final answer! A=85(pi) square units

[@travisloewen6191]

I never thought to use coordinates and the equation of a circle! I was able to solve it by using the chords of a circle theorem. I knew we had a chord with length 4 on the right. If we extend the vertical line on the left, we get another chord with length 12. The diameter is a perpendicular bisector of both chords. From there I used some Pythagorean theorem and got the same answer. I love that some of these problems can be solved in multiple ways. Thanks for sharing!

[@isting4741]

I think it's not so obvious to tell that the second chord has a lenght of 12 Instead you can draw a second chord from bottom left to top right point. The middle of the chord will be in the bottom left corner of a middle square and the perpendicular to that chord will intersect top side of that square in the middle. From that point it's quite simple to calculate radius using Pythagorean theorem.

[@MrJJbleeker]

I dont know why, but this one was very satisfying to watch 🤩

[@aryaaayra]

You never fail to impress me. I am envious.

[@gerardacronin334]

How spicy! 🌶️

[@SuspenduAuGaffa]

Loved this one, Andy. Ingenious solution! How spicy.

[@veryrandom6360]

Spiciest problem i've seen. Actually saw it a long time ago but had no idea on how to solve it. This is, to say the least, an amazing relief and very exciting to know now how it was solved

[@obelix8477]

That's a spicy assignment.

[@rukiyederenkoktepe2439]

90 degrees on the circle sees the diameter. We have two squares on the circle, we can draw two triangles with their hypothenuse equal to the diameter of the circle. Center of the circle is where two hypothenuses cross. Then basically you can find the unknown lenght by using similar triangle and pythagorean theorem. I like to solve every geometric question using triangles/rules cause thats basically all you need.

[@UsrYtb-zx3ru]

Soo cool ! I almost forgot how fun can it be.... Thanks for the great video!

[@numbers93]

You can also solve it by using the 3 points as vertices of a triangle. The circle is then the circumcircle of the triangle. The perpendicular bisectors come out nicely once we assign coordinates to these points. The slopes of these perpendicular bisectors are quite nice, and their intersection (the center of the circle) is an integer point just like the the vertices of the triangle. So then the square of the radius can be obtained using Pythagorean theorem. I used the top-right vertex and found 85 = 9^2 + 2^2.

[@dpeastman]

That's the way I did it. A bit easier I think (at least in this case where the numbers are nice and neat).

[@trickyricky1971]

New subscriber here, very spicy work! Im a CAD//CAM guy and circles are my life. I always enjoy a good circle riddle.

[@MarkTeng-hs5xh]

I’d like to share another geometric approach to this question. By connecting the three red dots you get a triangle which the circle is its circumcircle And simply by calculating the area and sides fo the triangle gives you the radius of the circumcircle Easier to formulate and calculate AT = area of triangle R = radius of the circle (a,b,c) = three sides of the triangle (A,B,C)=the three angles of the triangle for 0.5*ab*sinC = AT & 2*R*sinC = c We have (abc)/(4*AT) = R = sqrt(85) hence area of circle is 85*pi

[@garydunken7934]

That was exciting indeed. I spent few minutes thinking about how to solve it, but couldn't figure it out. Thanks for the video.

[@theporkchopexpress3939]

"How. Exciting." Gets me every time

[@Lattrodon]

I've been out of school for 8 years and I'm somehow interested in watching this math video

[@atrus3823]

This one took me a while. I took a totally different approach, which was very simple in setup, but took quite a bit of working out. I used trig. It's kind of hard to explain, but basically, using the coordinate system you use (though the way I did it, I didn't need a coordinate system), I'm looking for x, the distance from (0, 2) to the center. I can also find alpha, the angle between (0, 2) to (0, -14) and (0, 2) to (4, -14). I also note the unknown angle theta, the angle between the center to (0, 2) and the center to (0, 0). From these relationships, it's not too hard to see tan(alpha) = 1/4 and tan(theta) = 2/x. The last piece of the pie is noticing (through some angle relationships) that the angle between the center to (0, -14) and the center to (4, -14) is 2alpha + theta. This gives tan(2alpha + theta) = 6/(16 - x). I then used the tan sum and tan double angle identities to get the last equation in terms of tan(alpha), tan(theta), and x. Substituting, I can get an equation totally in terms of x. After a lot of algebra, I got 0 = 8x^2 - 8x - 576. Using our old friend the quadratic formula, there are two solutions: -8 and 9. It's obviously not -8, since it's a length, so it's 9! From that, I could get the radius with Pythagorus: 9^2 + 2^2 = r^2.

[@SogehtMathe]

2:34 Here I would just use the square root: We'll get: k = 4 - k | + k 2k = 4 | : 2 k = 2 Great video! Very well explained. 👌

[@silk.bonnet]

Phenomenal!! This really was a fun one

[@shteebo]

I'd never heard the term "equation of a circle", though seeing it, I imagine I could have derived it from basic knowledge of circles and coordinate geometry. This is certainly a perfect application of it. I love your style--no fluff, just a clear illustration quickly explained with work shown. I recommend your channel to everyone I meet who enjoys math.

[@DaSquyd]

What did you learn that it was called in school if not "equation of a circle"?

[@1959mikel]

Equation of a circle helps define a circle on the x, y co-ordinate system.

[@aperinich]

@@DaSquyd many secondary students would never touch upon this topic area.

[@bighammer3464]

How the heck did you take and pass geometry and not learn the equation of a circle yet you could “derive” it from “basic” knowledge 😂😂😂😂😂😂😂

[@aperinich]

@@bighammer3464 In high school you don't "take geometry" it's just part of a syllabus. Circle equation does not come up in any secondary basic level. which is the highest most people ever do.. You can however derive it from basic knowledge of numbers. Two variables, each squared, whose sum equals a constant, forms a circle. It is kind of obvious.

[@TingleCowboy]

Because two points of the right square touch the circle, the center of the circle is at the same height as the center of the square. You can then find the radius using two simple equations based on the Pythagorean theorem. X is the distance from the center of the circle to the right side of the right square: 2^2 + x^2 = r^2 (16-x)^2 + 6^2 = r^2 4 + x^2 = (16 - x)(16-x) + 36 x^2 = 256 - 16x - 16x + x^2 + 36 - 4 0 = 288 - 32x x = 288 / 32 ‎ = 9 r^2 = 9^2 + 4 ‎ = 85

[@mitchellclark4377]

You can also figure out the y-coordinate of the circle centre by symmetry, since the rightmost square is touching the circle at two points. I'm not sure how you would prove it, but that must mean the horizontal line bisecting the circle must be exactly equidistant to each of the two points, and since the side length is 4, the centre y-coord must be 2.

[@vbprogrammer95]

The moment he put it on the coordinate plane I knew how to solve it. Very clever because it makes it incredibly easy. Good job!

[@cremath]

That gives good vibes dude😅❤

[@chewiebacka4377]

The graphics you use are so easy to follow. Well done.

[@ShivangGulati]

"How Exciting" gets me everytime 😭😂

[@tomaszmazurek64]

Nice. Afaik there is a formula for this and I expected you to use it, but solving it using the circle equation was way more intuitive.

[@MatejVelican]

Oh how the universe aligns! Idk if you will believe me, but I am a drywaller. On monday I have to make a decorative circle in a ceiling and shit needs to hit few specific spots. This video is exactly what I will be using. Thanks yt recommendations!

[@jacobstump4414]

Question for you. At around 2:40, you have k^2 = (4-k)^2 Can you take the square root of both sides in this situation? The results still comes out to K = 2 but I seem to remember that there are some situations where you can take the square root of both sides of an equation, but there are other situations where you cannot. Is that right?

[@mr_mathur0925]

I have not yet studied the equation of circle but due to your awesome & brilliant explaination I have grasped the concept.

[@AbhishekJain-px1pm]

@AndyMath - How do you create these spicy solutions in the form of Animation? What software are you using? Would anyone else be able to help me?

[@konradyearwood5845]

Very good. Enjoyed the use of the circle formula.

[@popeyeschicken-u4f]

I had not thought of using analytic geometry to solve this, I instead went with the inscribed angle law to find the central angle between the two lower points then applied the cosine law to find the radius of the circle. This is one of my favorite things about math though, how completely different paths can lead to the same answer. It's always very interesting to see how other people think when tackling the same problem :)

[@AnweshAdhikari]

The solution brought a smile on my face! Not exactly the solution, but the approach!

[@Xyour_starX]

This is the only way to teach math. You are an artist

[@anjajohansson258]

I don't recall seeing this approach before. Very neat! I approve.

[@HolySoliDeoGloria]

Spicy good problem! I solved it in almost the same way without any peeks or hints-not even seeing the description that gave the hint of using a coordinate system; I set the lower left point at (0, 0). You can skip some algebra by recognizing that the horizontal diameter of the circle must pass through the point midway between the two points on the far right. In my coordinate system, that makes the y-coordinate of the center of the circle (k) equal to 6. In your coordinate system, k would be immediately recognized as 2. Then you can proceed to solve for h and then r^2.

[@albertosaldanha]

It's amazing! I'm a Math teacher in Brazil and I really love your videos

[@JoseAntonio-ng5yu]

There is a solution which I consider simpler through a bit hard to explain in a comment, but I'll try. Draw two chords: one from the left point to the top right point, the other a horizontal line to the bottom right point. The intersection of the chords is the lower right vortex of the second square. Use the chord theorem to compute the missing distance. If you draw two perpendicular lines that bisect the two chords, the intersection of such lines is the center of the circle. Through Pithagoras you can easily obtain d, the distance from the center to the intersection of the chords. Chord theorem then tells you r^2-d^2=80, which is the product of the chord sections. Sounds kind of hard, but it's actually quite easy

[@SoumyaDeyOnGoogle]

Which right angled triangle did you use to find d?

[@SoumyaDeyOnGoogle]

There is a typo in your comment. r^2 - d^2 = 60. Not 80. And d = 5.

[@JoseAntonio-ng5yu]

It's 80, because in the right triangle the sides are 8 and 4, so the hypothenuse is square root of 80. d is square root of five

[@Bokery1000]

Nice one! I've totally my lesson in analytical geometry. Thanks for reminding it. Very interesting solution

[@spencerhansen5595]

My first thought was just to draw perpendicular lins from the center of 2 of the chords defined by the points and calculate the distance between the point at which they cross and any of the 3 initial points

[@Michiel_de_Jong]

When you see that the midpoint must be on the perpendicular bisector of the two points of contact on the right... you can easily construct two triangles with only one variable 'x' to identify the midpoint and create one simple equation with Pythagoras. f.i. (10-x)^2 + 2^2 = (6+x)^2 + 6^6 . x^2 drops out and x=1 in this version. Both hypotenuse are √84 . Thus 84π

[@teej143]

Fantastic and making it look/sound easy

[@gregthomson1064]

Very well done and well explained! I forgot about the equation of a circle. I thought we were building triangles and going nuts on trig.

[@flikkie72]

I did it slightly differently. I do start with putting it on the coordinate plane, but then i try to find the center of the circle. For this, take the average of two combinations of the coordinates. Since one of the coordinates is set to 0,0, thats the easiest one to average the other 2 wirh. Those averages are: (-8, -2) and (0,2). If you imagine a line that goes through the "parent" coordinates and the average (child) coordinate, you can imagine a line that's perpendicular to that line going through the average point. Because one of the coordinates is the origin, the slope of that line is the negative inverse of the average coordinate, i.e. the perpendicular line through (-8, -2) has a slope of -8/2 = -4 (y per x). The other line has a slope of -0/2 = 0 (y per x). Therefore the y coordinate of the center of the circles is on the line y=2. If we set the equation for the other line we found equal to that y value, we get -2+(-4(x+8))=2, -4x-32=4, 36=4x, the center point is then (-9,2). If you apply Pythagoras to get the distance between this point and the origin (aka the radius(, you get √(9^9±2^2) = √85. The surface is then the square of that number time pi.

[@Gamesmarts194]

Huh, forgot that you could use a graph to solve that, that’s pretty neat! One thing I thought about was at 2:39 (solving [k^2=(4-k)^2] ), couldn’t you take the square root of both sides to get [k=4-k] (which then you add [k] to both sides to get [k+k=4] -> [2k=4] then divide both sides by [2] to get [k=2]) so you wouldn’t have to expand out [(4-k)^2]?

[@mr.d8747]

*Here's how I would solve it: connect the three points to get a triangle of base 4 and a height of 16. The other two sidelengths can be calculated by the Pythagorean Theorem. From there, the area of the triangle 1/2 • base • height. But another formula for the area of a triangle is the product of the sidelengts divided by the radius of the exocircle. Substituting in the values of the area and the sidelengths gives us the radius and from there the area is just r²π.*

[@puzzlesometime341]

Draw a line bisects the 4-length chord, then that line is diameter of the circle. From left bottom square, draw a vertical 12-length chord. Using intersecting chords theorem: diameter to 4-lengh chord and diameter to 12-length chord, calculating d = 2√85 => r = √85 => Area = 85π

[@karammohamed5569]

I did this defining the bottom left point on the bottom left square as the origin, finding the equations of the lines from that point to the other 2 points on the circumference, then finding the equations of 2 lines perpendicular to those 2 lines that pass through the midpoint (basically the perpendicular bisectors), then by finding the intersection point (7,6) i can calculate the radius by calculating the distance to the origin, since the origin is on the circumference of the circle (pythagoras) and then use that to calculate the area. Very exciting

[@buomar9129]

I would’ve never thought about using the coordinates for this problem. You’re brilliant! Please keep up making those short and intuitive problems.

[@jperez7893]

this is how solutions should be done. you didn't skip a step. students hate math because teachers skips steps. keep up the good work

[@jonathanmartin4919]

GENIUS I’m over here thinking how we can use arcs and angles to try and figure out the area. It never even occurred to me to plot the circle on a graph

[@thecodingfox]

Before watching the video, I saw "coordinate geometry" in the description, which gave me a hint to use the equation for a circle. Quite a spicy trick, I loved it! Keep making more geometry videos Andy 🏐📏📐👑

[@ercsrc]

It's much simpler without the coordinates. Two perpendicular triangles with hypothenus being the r 1- from center to bottom left corner on the circle 2- from center to right bottom corner on the circle. 6-7-sqrt(85) and 2-9-sqrt(85)

[@rogerlie4176]

Cosine + Pythagoras theorems gave me x^2 + 12x + 72 = r^2 x^2 -20x + 104 = r^2 where x is the distance from the intersection between the diagonal line from the leftmost quadrat and the horizontal line through the centre of the rightmost quadrat, and the circle's centre. Subtracting these equations shows that x = 1 and thus r^2 = 85.

[@olerask2457]

Nice solve. If the area of each square is generalized using A instead of 16, we get r^2 = 85*A/16.

[@miojos4n]

I'm in uni to study design but this guy's videos are so fun to watch

[@Imabearlol_]

Whoa, I did this a completely different way by using the inscribed angle theorem to compute the arc angle between the two points on the right, then using that to compute the radius of the circle. I think you got a cleaner form than me: (4pi)/(sin^2(arctan(1/2)-arctan(1/4)) isn't quite as clean as 85pi, but they're equal!

[@Trollllium]

I used the law of sines to solve it! View the three points touching the circle as vertices to a triangle. Then the circle is the circumcircle of the triangle. Let A be the top right vertex, B be bottom right, and C be bottom left. By law of sines, a/sin(A)=2R where R is the radius of the circumcircle and a=length of BC. sin(A) is easy to calculate: imagine a right triangle by extending AB downwards by another 4 units to point D. Then ADC is a right triangle, and we get AD=8, DC=16, so by Pythagorean AC=8sqrt(5), so sin(A)=opp/hyp=DC/AC=2/sqrt(5). Now we need a=BC, but this can be solved by applying Pythagorean on triangle BDC, since BD=4, DC=16, we have a=BC=4sqrt(17). Combining all these information together, we have R=a/2sin(A)=4sqrt(17)/(4/sqrt(5))=sqrt(85). The area of the circle is this πR^2=85π. Using this method I was able to mental math it in a few minutes 🤓

[@duongquocthongho2117]

you can also do R=abc/4S

[@steelcitytbirds]

I don’t recall being taught the formula for a circle. Well done!

[@johnspathonis1078]

Hi Andy Another much shorter method is to use intersecting chords. Once all the chord lengths are known use 4.R^2 = A^2 + B^2 + C^2 + D^2 where A, B , C ,D are the chord lengths.

[@semicolumnn]

Very nice. My idea was to draw the triangle formed by those points (up to scaling by 4 they have lengths 1, sqrt(17) and sqrt(20).) And then to find R (the circumradius) we can do a/sin A. We can find sin A (where a is the side of length 1) by finding the area bc /2 *sin A = 2. So sin A = 4/sqrt(340), 2R = sqrt(340)/4 = sqrt(85)/2 which means R^2 = 85/16. Rescaling by 4 we get 85pi as desired.

[@semicolumnn]

For the relevant search terms: This is by Pythagoras (to get the lengths), the law of sines (to find R) and the formula for the area of a triangle given one angle and two of its adjacent sides.

[@semicolumnn]

Apparently theres a neat formula for the area of a triangle that uses the law of sines to get [ABC] = abc/4R

[@LFTRnow]

That was a spicy application of co-ords to solve this problem. I was realizing you need to solve for r (or r^2) but couldn't figure out how to start. Co-ords to the rescue.

[@miscman84]

This is going to help me so much with my algebra EOC on Thursday, Thanks! That was spicy!

[@v1e1r1g1e1]

Yeah. Just got to know the grid formula from the get-go, right? Any other ways of doing this WITHOUT the formula?

[@dominicwalker1899]

So genius ❤ love it!

[@RinusKoe-l2d]

@2:40 for k^2 = (4-k)^2, is there a reason the right hand side needs to be expanded first? what i did is just square root both sides to get k = 4-k, k=2 is there a potential problem with this?

[@micahlong2073]

I would have taken the perpendicular lines from the lines 1) between the two corners of the "double-touch" square and 2) between the bottom of the "double-touch" corner and the "single-touch" square. These are chords so their perpendicular bisectors (which you can solve for) will intersect at the centre of the circle.

[@erekte7266]

Easier solution is using 3 points and finding angle on the circle. Than double it for the angle from the center and use it to find area of the circle. Working on angular system is usually much easier than cartesian system.