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Writing standard equation of a circle

Given a circle on the coordinate plane, Sal finds its standard equation, which is an equation in the form (x-a)²+(y-b)²=r².

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  • duskpin ultimate style avatar for user Vanshika
    Around , he explains that we need to use the pythagorean theory to find the radius r. But can't we just estimate the no. of units from the centre (point -1, 1) to the point (7.5, 1), which also lies on the circumference?
    (9 votes)
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  • orange juice squid orange style avatar for user Andrea Mitchell
    How do you put the square root of (#-#)squared + (#-#)squared, into a calculator? Or how do you solve this without a calculator?
    (9 votes)
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    • piceratops ultimate style avatar for user Christinabug
      Some calculators are really simple and can't find exponents or square roots, but others can. You make a square root symbol, and then fit all of (#-#)^2 + (#-#)^2 inside of it. If you don't have a calculator that can do that, I guess you can just do it out with pencil and paper. Subtract #-# inside the parentheses, then use long multiplication to "square" the result (just multiply by itself), and then.... I think there's a way to find square roots on paper, but you might have to find an online calculator.
      (13 votes)
  • duskpin ultimate style avatar for user FemiO
    At , He said that -5 squared is 25. Isn't it -25?
    (2 votes)
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  • duskpin tree style avatar for user gcordner04
    At , Sal makes a right triangle, do you have to make a right triangle, can't you just measure straight up, down, to the right or to the left to find the radius because every point on a circle is the same distance from the center?
    (5 votes)
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  • blobby green style avatar for user Cynthia Frank
    I get so confused in this really helps but I still don't understand can someone please help?
    (2 votes)
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    • female robot grace style avatar for user loumast17
      First you need to know that the equation for a circle is (x-a)^2 + (y-b)^2 = r^2 where the center is at point (a,b) and the radius is r. so for instance (x-2)^2 + (y-3)^2 = 4 would have the center at (2,3) and have a radius of 2 since 4 = 2^2. This means if you went a distance of 2 away from that center point you would be on the circle. Some super simple ones are adding 2 to the x or y coordinate. so the center at (2,3) means (2+2, 3), (2-2, 3), (2, 3+2) and (2, 3-2) are on the circle.

      Now the video starts giving a circle, where we know one point on the circle. Thankfully it tells us the center, so we can handle the first half of the equation. (x-a)^2 + (y-b)^2 where a = -1 ad b = 1 makes it (x+1)^2 + (b-1)^2. So easy. The radius isn't so easy.

      there are no points directly above, below or to the left or right of the center you can gauge the radius from. In fact you only know for sure of one point since it tells you. Well, the radius is the distance from the center of the circle to the edge, and the other point we know is by defiition on the edge. Now, how do you find the distance between two points.

      If you imagine the two points had a horizontal and vertical line connecting them, this would make a right triangle. Can you manage from there?
      (4 votes)
  • aqualine ultimate style avatar for user Trixia Aster Dado
    what if the given were the endpoints of a diameter? how can you solve it?
    (2 votes)
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    • mr pink red style avatar for user andrewp18
      Use the distance formula to find the length of the diameter, and then divide by 2 to get the radius. Then find the midpoint of the diameter which will be the center of the circle. Now you have the coordinates of the center and the radius and that is all that is necessary to write the standard equation of the circle.
      (5 votes)
  • boggle blue style avatar for user JACOB'S SUNG
    The equation gets 74 at the end, can someone tell me what 74 is?
    (3 votes)
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  • duskpin ultimate style avatar for user leahm
    Hi! Instead of using pythagorean theorem, can we use the distance formula? Isn't that plausible? Thanks
    (2 votes)
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  • piceratops ultimate style avatar for user ∫∫∫ Hannan dG dB dR
    How does sqrt(1 + 0.75^2) equal 1.25? I'm really stuck
    (1 vote)
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  • blobby green style avatar for user Kian Uduman
    What if I don't have a center, or a radius, but 4 lines, x=7, y=-2, x=-1, and y=-10?
    (1 vote)
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    • stelly blue style avatar for user Kim Seidel
      As @kubleeka indicated, your lines create a square. Were you asked to create a circle that fits within the square? Are these suppose to be the outer limits for the circle?

      If so, you can find the center of the circle by averaging the X-values to get its X-coordinate. And, repeat the process for the Y's to get the Y-coordinate.

      To find the radius: Check the distance between the lines and split the distance in half.
      X: [7-(-1)]/2 = 8/2 = 4
      Y: [-2-(-1)]/2=8/2 = 4
      Your radius = 4

      From this info you can now write an equation for a circle that sits within the lines.

      Hope this helps.
      (3 votes)

Video transcript

- [Voiceover] So we have a circle here and they specified some points for us. This little orangeish, or, I guess, maroonish-red point right over here is the center of the circle, and then this blue point is a point that happens to sit on the circle. And so with that information, I want you to pause the video and see if you can figure out the equation for this circle. Alright, let's work through this together. So let's first think about the center of the circle. And the center of the circle is just going to be the coordinates of that point. So, the x-coordinate is negative one and then the y-coordinate is one. So center is negative one comma one. And now, let's think about what the radius of the circle is. Well, the radius is going to be the distance between the center and any point on the circle. So, for example, for example, this distance. The distance of that line. Let's see I can do it thicker. A thicker version of that. This line, right over there. Something strange about my... Something strange about my pen tool. It's making that very thin. Let me do it one more time. Okay, that's better. (laughs) The distance of that line right over there, that is going to be the radius. So how can we figure that out? Well, we can set up a right triangle and essentially use the distance formula which comes from the Pythagorean Theorem. To figure out the length of that line, so this is the radius, we could figure out a change in x. So, if we look at our change in x right over here. Our change in x as we go from the center to this point. So this is our change in x. And then we could say that this is our change in y. That right over there is our change in y. And so our change in x-squared plus our change in y-squared is going to be our radius squared. That comes straight out of the Pythagorean Theorem. This is a right triangle. And so we can say that r-squared is going to be equal to our change in x-squared plus our change in y-squared. Plus our change in y-squared. Now, what is our change in x-squared? Or, what is our change in x going to be? Our change in x is going to be equal to, well, when we go from the radius to this point over here, our x goes from negative one to six. So you can view it as our ending x minus our starting x. So negative one minus negative, sorry, six minus negative one is equal to seven. So, let me... So, we have our change in x, this right over here, is equal to seven. If we viewed this as the start point and this as the end point, it would be negative seven, but we really care about the absolute value in the change of x, and once you square it it all becomes a positive anyway. So our change in x right over here is going to be positive seven. And our change in y, well, we are starting at, we are starting at y is equal to one and we are going to y is equal to negative four. So it would be negative four minus one which is equal to negative five. And so our change in y is negative five. You can view this distance right over here as the absolute value of our change in y, which of course would be the absolute value of five. But once you square it, it doesn't matter. The negative sign goes away. And so, this is going to simplify to seven squared, change in x-squared, is 49. Change in y-squared, negative five squared, is 25. So we get r-squared, we get r-squared is equal to 49 plus 25. So what's 49 plus 25? Let's see, that's going to be 54, or was it 74. r-squared is equal to 74. Did I do that right? Yep, 74. And so now we can write the equation for the circle. The circle is going to be all of the points that are, well, in fact, let me right all of the, so if r-squared is equal to 74, r is equal to the square-root of 74. And so the equation of the circle is going to be all points x comma y that are this far away from the center. And so what are those points going to be? Well, the distance is going to be x minus the x-coordinate of the center. x minus negative one squared. Let me do that in a blue color. Minus negative one squared. Plus y minus, y minus the y-coordinate of the center. y minus one squared. Squared. Is equal, is going to be equal to r-squared. Is going to be equal to the length of the radius squared. Well, r-squared we already know is going to be 74. 74. And then if we want to simplify it a little bit, you subtract a negative, this becomes a positive. So it simplifies to x plus one squared plus y minus one squared is equal to 74. Is equal to 74. And, we are all done.