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Integrated math 2

Course: Integrated math 2 > Unit 12

Lesson 4: Expanded equation of a circle
Math
Integrated math 2
Conic sections
Expanded equation of a circle
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Circle equation review

Google Classroom
Review the standard and expanded forms of circle equations, and solve problems concerning them.

What is the standard equation of a circle?

(x−h)2+(y−k)2=r2
This is the general standard equation for the circle centered at (h,k) with radius r.
Circles can also be given in expanded form, which is simply the result of expanding the binomial squares in the standard form and combining like terms.
For example, the equation of the circle centered at (1,2) with radius 3 is (x−1)2+(y−2)2=32. This is its expanded equation:
(x−1)2+(y−2)2=32(x2−2x+1)+(y2−4y+4)=9x2+y2−2x−4y−4=0
Want to learn more about circle equations? Check out this video.

Practice set 1: Using the standard equation of circles

Problem 1.1
(x+4)2+(y−6)2=48
What is the center of the circle?
(
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi
,
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi
)
What is its radius?
If necessary, round your answer to two decimal places.
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi
units

Want to try more problems like this? Check out this exercise and this exercise.

Practice set 2: Writing circle equations

Problem 2.1
A circle has a radius of 13 units and is centered at (−9.3,4.1).
Write the equation of this circle.

Want to try more problems like this? Check out this exercise.

Practice set 3: Using the expanded equation of circles

To interpret the expanded equation of a circle, we should rewrite it in standard form using the method of "completing the square."
Consider, for example, the process of rewriting the expanded equation x2+y2+18x+14y+105=0 in standard form:
x2+y2+18x+14y+105=0x2+y2+18x+14y=−105(x2+18x)+(y2+14y)=−105(x2+18x+81)+(y2+14y+49)=−105+81+49(x+9)2+(y+7)2=25(x−(−9))2+(y−(−7))2=52
Now we can tell that the center of the circle is (−9,−7) and the radius is 5.
Problem 3.1
x2+y2−10x−16y+53=0
What is the center of this circle?
(
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi
,
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi
)
What is the radius of this circle?
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi
units

Want to try more problems like this? Check out this exercise and this exercise.

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  • aqualine tree style avatar for user Ravengal101
    Posted 7 years ago. Direct link to Ravengal101's post “What should be done to ad...”
    What should be done to adjust the radius of the circle in the practice? I could move the centre, but I'm lost as to how the radius should be adjusted.
    (25 votes)
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  • piceratops ultimate style avatar for user M. McManus
    Posted 6 years ago. Direct link to M. McManus's post “Parabolas are called Quad...”
    Parabolas are called Quadratics. What are circle equations called? Are they also called Quadratics because of the square? Thanks.
    (8 votes)
    Default Khan Academy avatar avatar for user
    • leaf green style avatar for user kubleeka
      Posted 6 years ago. Direct link to kubleeka's post “Parabolas, hyperbolas, el...”
      Great Answer
      Good Answer
      Parabolas, hyperbolas, ellipses, and circles (which are just special ellipses) are collectively called "Conic Sections", since you can find each of these curves as the intersection between an infinite cone and a plane.

      "Quadratic" refers to a polynomial of degree 2, and hence only describes parabolas.

      Equations that describe circles don't really have a special name.
      (27 votes)
  • blobby green style avatar for user 1262076
    Posted 7 years ago. Direct link to 1262076's post “In problem 2.2 how did yo...”
    In problem 2.2 how did you get √17?
    (6 votes)
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    • aqualine tree style avatar for user Judith Gibson
      Posted 7 years ago. Direct link to Judith Gibson's post “I think the graph is dece...”
      Good Answer
      I think the graph is deceiving in that it is hard to read accurately.
      At first glance it looks like the circle goes though the point (9,0) and therefore has a radius of 4.
      However, upon closer looking, the circumference of the circle is just a bit beyond (9,0) and actually hits the point (9,-1).
      That's why (using the Pythagorean theorem) the distance for the radius is calculated between the center (5,0) and the point (9,-1) which lies on its circumference. (Note that the point (9,1) could have been used instead as could (1,1) or (1,-1).)
      (22 votes)
  • blobby green style avatar for user arthurgrayuk
    Posted 4 years ago. Direct link to arthurgrayuk's post “How can you find square r...”
    How can you find square roots without a calculator?
    (4 votes)
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    • starky ultimate style avatar for user KLaudano
      Posted 4 years ago. Direct link to KLaudano's post “If you don't need an exac...”
      If you don't need an exact answer, you can use linear approximations. Suppose we are trying to find the square root of a number, x, and we know the square root of a number, k, that is "close" to x. (k will likely be perfect square as it makes the math easier). The square root of x is approximately equal to (x + k)/(2 * sqrt(k)).
      (8 votes)
  • blobby green style avatar for user wairishtinah
    Posted 6 years ago. Direct link to wairishtinah's post “Find the centre and radiu...”
    Find the centre and radius of a circle whose equation is x2+y2-4y-21
    (5 votes)
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    • aqualine tree style avatar for user Judith Gibson
      Posted 6 years ago. Direct link to Judith Gibson's post “What you have written isn...”
      What you have written isn't an equation because it has no equals sign, so presumably you mean x^2 + y^2 - 4y - 21 = 0
      We know that the general equation for a circle is ( x - h )^2 + ( y - k )^2 = r^2, where ( h, k ) is the center and r is the radius.
      So add 21 to both sides to get the constant term to the righthand side of the equation.
      x^2 + y^2 -4y = 21
      Then complete the square for the y terms.
      x^2 + y^2 - 4y + 4 = 21 + 4
      Then factor.
      x^2 + ( y - 2 )^2 = 5^2
      So, the center is at ( 0, 2 ) and the radius is 5.
      (4 votes)
  • blobby green style avatar for user Louise
    Posted 6 years ago. Direct link to Louise's post “Hi! I really need an urge...”
    Hi! I really need an urgent response towards how should I find the equation of circle if the only given were the center (-1,-7) and a point (2,11).

    Well, I tried doing using the standard equation of the circle and imputed the given sample and tried to square it. Unfortunately, WileyPLUS (Its a book converted into sort of online stuffs and other than pen and paper activities its all made online) as part of our school marked my answers wrong... can anyone help me? I really need to perfect it.
    (2 votes)
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    • aqualine tree style avatar for user Judith Gibson
      Posted 6 years ago. Direct link to Judith Gibson's post “The standard equation for...”
      Good Answer
      The standard equation for a circle centred at (h,k) with radius r
      is (x-h)^2 + (y-k)^2 = r^2
      So your equation starts as ( x + 1 )^2 + ( y + 7 )^2 = r^2
      Next, substitute the values of the given point (2 for x and 11 for y), getting
      3^2 + 18^2 = r^2,
      so r^2 = 333.
      The final equation is (x+1)^2 + (y+7)^2 = 333
      Hope this helps!
      (11 votes)
  • blobby green style avatar for user Ann Meneses
    Posted 6 years ago. Direct link to Ann Meneses's post “What if the equation is 9...”
    What if the equation is 9x^2 + 24xy + 16y^2 + 90x - 130y = 0 ? I can't figure this out. It says that I should rotate the axes to remove the xy, but I can't really get it.
    (4 votes)
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    • piceratops seed style avatar for user Chris O'Donnell
      Posted 6 years ago. Direct link to Chris O'Donnell's post “A=9, B=24, C=16 so the di...”
      A=9, B=24, C=16 so the discriminant is 576 - 4(9)(16) = 576 - 576 = 0. Therefore this is a parabola. Factorising the first three terms we get (3x+4y)^2 + 90x - 130y = 0. Let u = 3x+4y and v = -4x+3y. Then x = (3u - 4v)/25 and y = (4u + 3v)/25. So the equation becomes:

      u^2 + (18/5) (3u-4v) + (26/5) (4u+3v) = u^2 + 158u/5 + 6v/5 = 0

      Complete the square: (u+79/5)^2 - 6241/25 + 6v/5 = 0

      Rearrange for v: v = 6241/30 - 5/6 (u+79/5)^2

      So the directrix is v = 6241/30 + 6/20 = 625/3
      And the focus is (u,v) = (-79/5, 6241/30 - 6/20) = (-79/5, 3116/15)

      Writing in terms of x and y:
      The directrix is -12x+9y = 625
      And the focus is (x,y) = (-527/15, 112/5)
      (3 votes)
  • blobby green style avatar for user sheeraz.mohammd
    Posted 3 years ago. Direct link to sheeraz.mohammd's post “why does the formula for ...”
    why does the formula for the equation of a circle give you centre?
    like (x+3)^2+(y+3)^2 = 25
    why is the centre -3,3, when plugged in, it clearly does not equal 25
    (2 votes)
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    • stelly blue style avatar for user Kim Seidel
      Posted 3 years ago. Direct link to Kim Seidel's post “You are looking at it in ...”
      You are looking at it in the wrong way. The formula for a circle is based upon the distance formula for finding the distance between 2 points. In this case one of the points is the center point (-3,3) and you are finding all points that have a distance of 5 units from the center point.
      Hope this helps.
      (6 votes)
  • blobby green style avatar for user Bhoke Maseke
    Posted 2 years ago. Direct link to Bhoke Maseke's post “i still dont get it at t...”
    i still dont get it at the practice exercise three
    (2 votes)
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    • mr pink green style avatar for user David Severin
      Posted 2 years ago. Direct link to David Severin's post “Exercise 3 requires you t...”
      Exercise 3 requires you to complete the square for x and y. So if you have x^2+y^2-10x-16x+53=0, first put variable terms together (x^2-10x+ __) + (y^2-16y+ ___) - ___ - _____ +53=0. To complete the square, you have to divide middle term by 2 and square the result, so -10/2 = -5, (-5)^2=25 and -16/2=-8, (-8_^2=64. Thus, (x^2-10x+25) + (y^2-16x+64)-25-64+53=0. So (x-5)^2 + (y-8)^2 - 36=0.
      (5 votes)
  • blobby green style avatar for user sugutjune
    Posted 5 years ago. Direct link to sugutjune's post “How do i expand to get 81...”
    How do i expand to get 81 and 49 highlighted? please guide me on the video that might help me more. cos i understand how you did it but i don't understand how to tackle that particular step
    (3 votes)
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