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source : mathsisfun.com

## Circle on a Graph

Let 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:

x
y
x2 + y2
5
0
52 + 02 = 25 + 0 = 25
3
4
32 + 42 = 9 + 16 = 25
0
5
02 + 52 = 0 + 25 = 25
−4
−3
(−4)2 + (−3)2 = 16 + 9 = 25
0
−5
02 + (−5)2 = 0 + 25 = 25

In all cases a point on the circle follows the rule x2 + y2 = radius2

We can use that idea to find a missing value

Example: x value of 2, and a radius of 5

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 Case

Now 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:

(x−a)2 + (y−b)2 = r2

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.

Example: A circle with center at (3,4) and a radius of 6:

(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.

## “General Form”

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

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 Form

Now 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:

Example: x2 + y2 − 2x − 4y − 4 = 0

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 Circle

If we place the circle center at (0,0) and set the radius to 1 we get:

(x−a)2 + (y−b)2 = r2

(x−0)2 + (y−0)2 = 12

x2 + y2 = 1

Which is the equation of the Unit Circle

## How to Plot a Circle by Hand

1. 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!

Example: Plot (x−4)2 + (y−2)2 = 25

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:

The Center: (4,2)
Up: (4,2+5) = (4,7)
Down: (4,2−5) = (4,−3)
Left: (4−5,2) = (−1,2)
Right: (4+5,2) = (9,2)

Now, just sketch in the circle the best we can!

## How to Plot a Circle on the Computer

We 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.

Example: Plot (x−4)2 + (y−2)2 = 25

So the center is at (4,2), and the radius is √25 = 5

Rearrange to get “y=”:

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” …there can be two square roots!)

Move the “−2” to the right:y = 2 ± √[25 − (x−4)2]

So when we plot these two equations we should have a circle:

y = 2 + √[25 − (x−4)2]
y = 2 − √[25 − (x−4)2]

Try plotting those functions on the Function Grapher.

It is also possible to use the Equation Grapher to do it all in one go.

Circle – Wikipedia – Projecting a sphere to a plane. Outline. History. Geometers. v. t. e. A circle is a shape consisting of all points in a plane that are at a given distance from a given point, the centre…Start studying Equation of a Circle. Learn vocabulary, terms and more with flashcards, games and other study tools. What is the radius of a circle whose equation is x2 + y2 + 8x – 6y + 21 = 0?State the radius and center of the circle with equation 25 = x2 + (y + 3)2. The numerical side tells me that r2 = 25, so r = 5. The x-squared part is really (x – 0)2, so h = 0. The temptation is to read off the "3" from the y-squared part and conclude that k is 3, but this is wrong. The center-vertex form has…

Equation of a Circle Flashcards | Quizlet – …with radius 5 whose centre lies on x-axis and passes through the point (2, 3). We know that equation of circle is (x – h)2 + (y – k)2 = r2 Centre of circle is denoted by (h, k) Since it (h, 0) & given Radius = 5 Distance between centre and point on circle = radius Distance between points (h, 0) & (2,3) = 5…The equation of a circle centered at point (x1,y1) with a radius r is given by[tex](x-x_1)^2 + (y-y_1)^2 = r^2[/tex]are you sure you wrote down the problem corr…8 time by 2 = 16 which is the diameter of the circle. 1 decade ago. The radius of this circle is 8, so its circumference is 16π. Source(s): I have taught math for over 40 yr.

Conics: Circles: Working with Equations – Learn how to find the equation of a circle and use the discriminant to prove for tangency in intersections for Higher Maths. This equation is the same as the general equation of a circle, it's just written in a different form.The general form of the equation for a circle with its center at (h, k) and radius of r is: Let's get your equation into this form, then we can find the center (h, k) and the radius Complete the square to get: x^2+6x+9 + y^2 = -8+9 Factor to get: (x+3)^2 + (y-0)^2 = 1 Center at (-3,0) Radius = 1 Cheers, Stan H.Explanation: Equation of a circle is x^2+y^2=r^2 r^2=9 r=3. Benito has a bamboo stick measuring 70 cm long.he cuts it into 3 pieces.the first half the length of the second piece,which is the longest?.if the lon … gest piece is 28.4 cm,how long is each of the other two pieces?