Show A circle is easy to make: Draw a curve that is "radius" away And so: All points are the same distance In fact the definition of a circle is Circle on a GraphLet us put a circle of radius 5 on a graph: Now let's work out exactly where all the points are. We make a right-angled triangle: And then use Pythagoras: x2 + y2 = 52 There are an infinite number of those points, here are some examples:
In all cases a point on the circle follows the rule x2 + y2 = radius2 We can use that idea to find a missing value
Start with:x2 + y2 = r2 Values we know:22 + y2 = 52 Rearrange: y2 = 52 − 22 Square root both sides: y = ±√(52 − 22) Solve:y = ±√21 y ≈ ±4.58... (The ± means there are two possible values: one with + the other with −) And here are the two points: More General CaseNow let us put the center at (a,b) So the circle is all the points (x,y) that are "r" away from the center (a,b). Now lets work out where the points are (using a right-angled triangle and Pythagoras): It is the same idea as before, but we need to subtract a and b: And that is the "Standard Form" for the equation of a circle! It shows all the important information at a glance: the center (a,b) and the radius r.
Start with: (x−a)2 + (y−b)2 = r2 Put in (a,b) and r: (x−3)2 + (y−4)2 = 62 We can then use our algebra skills to simplify and rearrange that equation, depending on what we need it for. Try it Yourselfimages/circle-equn.js But you may see a circle equation and not know it! Because it may not be in the neat "Standard Form" above. As an example, let us put some values to a, b and r and then expand it
Start with:(x−a)2 + (y−b)2 = r2 Example: a=1, b=2, r=3:(x−1)2 + (y−2)2 = 32 Expand: x2 − 2x + 1 + y2 − 4y + 4 = 9 Gather like terms:x2 + y2 − 2x − 4y + 1 + 4 − 9 = 0 And we end up with this: x2 + y2 − 2x − 4y − 4 = 0 It is a circle equation, but "in disguise"! So when you see something like that think "hmm ... that might be a circle!" In fact we can write it in "General Form" by putting constants instead of the numbers:
x2 + y2 + Ax + By + C = 0 Note: General Form always has x2 + y2 for the first two terms. Going From General Form to Standard FormNow imagine we have an equation in General Form: x2 + y2 + Ax + By + C = 0 How can we get it into Standard Form like this? (x−a)2 + (y−b)2 = r2 The answer is to Complete the Square (read about that) twice ... once for x and once for y:
Start with:x2 + y2 − 2x − 4y − 4 = 0 Put xs and ys together:(x2 − 2x) + (y2 − 4y) − 4 = 0 Constant on right:(x2 − 2x) + (y2 − 4y) = 4 Now complete the square for x (take half of the −2, square it, and add to both sides): (x2 − 2x + (−1)2) + (y2 − 4y) = 4 + (−1)2 And complete the square for y (take half of the −4, square it, and add to both sides): (x2 − 2x + (−1)2) + (y2 − 4y + (−2)2) = 4 + (−1)2 + (−2)2 Tidy up:
Simplify:(x2 − 2x + 1) + (y2 − 4y + 4) = 9 Finally:(x − 1)2 + (y − 2)2 = 32 And we have it in Standard Form! (Note: this used the a=1, b=2, r=3 example from before, so we got it right!) Unit CircleIf we place the circle center at (0,0) and set the radius to 1 we get:
How to Plot a Circle by Hand1. Plot the center (a,b) 2. Plot 4 points "radius" away from the center in the up, down, left and right direction 3. Sketch it in!
The formula for a circle is (x−a)2 + (y−b)2 = r2 So the center is at (4,2) And r2 is 25, so the radius is √25 = 5 So we can plot:
Now, just sketch in the circle the best we can! How to Plot a Circle on the ComputerWe need to rearrange the formula so we get "y=". We should end up with two equations (top and bottom of circle) that can then be plotted.
So the center is at (4,2), and the radius is √25 = 5 Rearrange to get "y=":
Start with: (x−4)2 + (y−2)2 = 25 Move (x−4)2 to the right: (y−2)2 = 25 − (x−4)2 Take the square root: (y−2) = ± √[25 − (x−4)2] (notice the ± "plus/minus" ... Move the "−2" to the right:y = 2 ± √[25 − (x−4)2] So when we plot these two equations we should have a circle:
Try plotting those functions on the Function Grapher. It is also possible to use the Equation Grapher to do it all in one go. 8526, 8527, 8539, 8540, 8515, 8516, 569, 8544, 8559, 8560, 570, 1209 Copyright © 2022 Rod Pierce
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In the standard coordinate plane, what is the radius and the center of the circle ?Possible Answers:
Correct answer: Explanation: When finding the center and radius of circle , the center is and the radius is . Notice that they are not negative even though in the equation they have negative signs in front. This becomes important when dealing with real numbers. Also, notice the square of .Our circle, has the same principles applied as the above principle, therefore is our center. Notice how the numbers signs have been switched. This is the case for all circles due to the negative in the base equation above.To find the radius of a circle, you must take the number the equation is equal to and square root it. This is due to the square of mentioned above. The . Use the least common multiples of 27 to find that three 3’s make up 27. Take two threes out as the square root of a number multiplied by itself is itself. This leaves one 3 under the radical. Therefore our radius is .Center: Radius:
In the standard coordinate plane, what is the area of the circle ?Possible Answers:
Correct answer: Explanation: The general equation of a circle is .According to the question, . Thus, .The general equation for the area of a circle is .When we plug in 13 for , we get our area to equal .
A circle in the standard coordinate plane is tangent to the x-axis at (3,0) and tangent to the y-axis at (0,3). What is the equation of the circle? Possible Answers:
Correct answer: Explanation: The formula for the equation of a circle is (x – h)2+ (y – k)2 = r2, where (h, k) represents the coordinates of the center of the circle, and r represents the radius of the circle. If a circle is tangent to the x-axis at (3,0), this means it touches the x-axis at that point. If a circle is tangent to the y-axis at (0,3), this means it touches the y-axis at that point. Given these two points, we can determine the center and the radius of the circle. The center of the circle must be equidistant from any of the points on the circumference. This means that both (0,3) and (3,0) are the same distance from the center. If we draw these points on a coodinate plane, it becomes apparent that the center of the circle must be (3,3). This point is exactly three units from each of the given points, indicating that the radius of the circle is 3. When we input this information into the formula for a circle, we get (x – 3)2 + (y – 3)2 = 9.
Find the equation of the circle with center coordinates of and a radius of .Possible Answers:
Correct answer: Explanation: The equation of a circle is The center is or, written another way . Substituting for and for , our formula becomesFinally, the formula of the circle is
On the xy plane, what is the area of a circle with the following equation: Possible Answers:
Correct answer: Explanation: The standard form equation of a circle is , where is the center of the circle and is equal to the radius. Thus, since we have the circle's standard form equation already given to us, we can ignore and , since all we need is .The area of circle is equal to , which is equal to .
A circle has a center at (5,5) and a radius of 2. If the format of the equation for the circle is (x-A)2+(y-B)2=C, what is C? Possible Answers:
Explanation: The circle has a center at (5,5) and a radius of 2. Therefore, the equation is (x-5)2+(y-5)2=22, or (x-5)2+(y-5)2=4.
If the center of a circle is at (0,4) and the diameter of the circle is 6, what is the equation of that circle? Possible Answers:
Correct answer: x2 + (y-4)2 = 9 Explanation: The formula for the equation of a circle is: (x-h) 2 + (y-k)2 = r2 Where (h,k) is the center of the circle. h = 0 and k = 4 and diameter = 6 therefore radius = 3 (x-0) 2 + (y-4)2 = 32 x2 + (y-4)2 = 9
Circle A is given by the equation (x – 4)2 + (y + 3)2 = 29. Circle A is shifted up five units and left by six units. Then, its radius is doubled. What is the new equation for circle A? Possible Answers:
(x – 10)2 + (y + 8)2 = 116
(x + 2)2 + (y – 2)2 = 116
(x – 10)2 + (y + 8)2 = 58
Correct answer: (x + 2)2 + (y – 2)2 = 116 Explanation: The general equation of a circle is (x – h)2 + (y – k)2 = r2, where (h, k) represents the location of the circle's center, and r represents the length of its radius. Circle A first has the equation of (x – 4)2 + (y + 3)2 = 29. This means that its center must be located at (4, –3), and its radius is √29. We are then told that circle A is shifted up five units and then left by six units. This means that the y-coordinate of the center would increase by five, and the x-coordinate of the center would decrease by 6. Thus, the new center would be located at (4 – 6, –3 + 5), or (–2, 2). We are then told that the radius of circle A is doubled, which means its new radius is 2√29. Now, that we have circle A's new center and radius, we can write its general equation using (x – h)2 + (y – k)2 = r2. (x – (–2))2 + (y – 2)2 = (2√29)2 = 22(√29)2 = 4(29) = 116. (x + 2)2 + (y – 2)2 = 116. The answer is (x + 2)2 + (y – 2)2 = 116.
Which of the following equations describes all the points (x, y) in a coordinate plane that are five units away from the point (–3, 6)? Possible Answers:
Correct answer: (x + 3)2 + (y – 6)2 = 25 Explanation: We are trying to find an equation for all of the points that are the same distance (5 units) from (–3, 6). The locus of all points equidistant from a single point is a circle. In other words, we need to find an equation of a circle. The center of the circle will be (–3, 6), and the radius, which is the distance from (–3,6), will be 5. The standard form of a circle is given below: (x – h)2 + (y – k)2 = r2, where the center is located at (h, k) and r is the length of the radius. In this case, h will be –3, k will be 6, and r will be 5. (x – (–3))2 + (y – 6)2 = 52 (x + 3)2 + (y – 6)2 = 25 The answer is (x + 3)2 + (y – 6)2 = 25.
What is the equation for a circle of radius 12, centered at the intersection of the two lines: y1 = 4x + 3 and y2 = 5x + 44? Possible Answers:
None of the other answers
(x - 22)2 + (y - 3)2 = 12
(x + 41)2 + (y + 161)2 = 144
(x - 41)2 + (y - 161)2 = 144
(x - 3)2 + (y - 44)2 = 144
Correct answer: (x + 41)2 + (y + 161)2 = 144 Explanation: To begin, let us determine the point of intersection of these two lines by setting the equations equal to each other: 4x + 3 = 5x + 44; 3 = x + 44; –41 = x To find the y-coordinate, substitute into one of the equations. Let's use y1: y = 4 * –41 + 3 = –164 + 3 = –161 The center of our circle is therefore: (–41, –161). Now, recall that the general form for a circle with center at (x0, y0) is: (x - x0)2 + (y - y0)2 = r2 For our data, this means that our equation is: (x + 41)2 + (y + 161)2 = 122 or (x + 41)2 + (y + 161)2 = 144
Alexia
University of Missouri-Columbia, Bachelor of Science, Biochemistry. Washington University in St Louis, Master of Science, Bio...
CaiJun
WuYi University, Bachelor of Economics, International Business.
Alexandra
The University of Texas at Dallas, Bachelors, Theatre Arts.
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