equations of circles algebra 2 homework

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Circle Equations

A circle is easy to make:

Draw a curve that is "radius" away from a central point.

All points are the same distance from the center.

In fact the definition of a circle is

Circle: The set of all points on a plane that are a fixed distance from a center.

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 :

x 2 + y 2 = 5 2

There are an infinite number of those points, here are some examples:

In all cases a point on the circle follows the rule x 2 + y 2 = radius 2

We can use that idea to find a missing value

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

(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 = r 2

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:

Start with:

Put in (a,b) and r:

(x−3) 2 + (y−4) 2 = 6 2

We can then use our algebra skills to simplify and rearrange that equation, depending on what we need it for.

Try it Yourself

"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

And we end up with this:

x 2 + y 2 − 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:

x 2 + y 2 + Ax + By + C = 0

Note: General Form always has x 2 + y 2 for the first two terms .

Going From General Form to Standard Form

Now imagine we have an equation in General Form :

How can we get it into Standard Form like this?

The answer is to Complete the Square (read about that) twice ... once for x and once for y :

Example: x 2 + y 2 − 2x − 4y − 4 = 0

Now complete the square for x (take half of the −2, square it, and add to both sides):

(x 2 − 2x + (−1) 2 ) + (y 2 − 4y) = 4 + (−1) 2

And complete the square for y (take half of the −4, square it, and add to both sides):

(x 2 − 2x + (−1) 2 ) + (y 2 − 4y + (−2) 2 ) = 4 + (−1) 2 + (−2) 2

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:

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 = r 2

So the center is at (4,2)

And r 2 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.

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

Rearrange to get "y=":

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.

Equation of Circle Worksheet (pdf)

Students will explore the equation of a circle . Specifically, students will practice writing the equation of a circle from a graph as card as graphing a circle given its equation .

Example Questions

1) Identify the coordinates of the center and the length of the radius in the circle below (x - 5) 2 + (y + 2) 2 = 4 radius: Center: (___, _____)

2) Graph a circle centered at (5, 1) with a radius of 5

Visual Aids

Other Details

This is a 3 part worksheet:

• Part I Model Problems
• Part II Practice Problems
• Part III Homework. Mixed questions involving the equation of a circle.
• Equation of Circle
• Interactive Equation of Circle : explore the relationship between the equation and graph of a circle by clicking and dragging..

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5.1: The Equation of the Circle

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• Page ID 40924

• Richard W. Beveridge
• Clatsop Community College

The equation for a circle is typically given as: \[ (x-h)^{2}+(y-k)^{2}=r^{2} \] In this equation, the point \((h, k)\) represents the center of the circle and \(r\) represents the radius of the circle. This equation is derived from the distance formula. The definition of a circle is the locus (or collection) of points that are equidistant from a given point (the center of the circle).

So, taking this definition, we can say that any point \((x, y)\) that is on the circle should be a distance of \(r\) from the center. Using the distance formula: \[ d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}} \] Squaring both sides: \[ d^{2}=\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2} \] In this situation, the distance \(d\) is the radius of the circle, \(r .\) This distance should be the same for all points on the circle. So any point on the edge of the circle, \((x, y)\) should be a distance of \(r\) from the center of the circle, \((h, k)\) \[ (x-h)^{2}+(y-k)^{2}=r^{2} \]

Exercises \(\PageIndex{1}\)

Determine the center and radius for each circle and sketch the graph of the circle. 1) \(\quad(x-3)^{2}+(y+1)^{2}=9\) 2) \(\quad(x+4)^{2}+(y+2)^{2}=16\) 3) \(\quad(x-1)^{2}+(y-6)^{2}=20\) 4) \(\quad(x-2)^{2}+(y+5)^{2}=27\) 5) \(\quad(x+7)^{2}+(y-4)^{2}=33\) 6) \(\quad(x-5)^{2}+(y-3)^{2}=50\)

Sometimes the equation for a circle is not given in the standard form. In this situation, you need to put the equation into standard form and then determine the center and radius. In order to put the equation into standard form you will need to complete the square. Completing the square is a mathematical technique that is often useful and is the basis for how the quadratic formula is derived.

Suppose that we are given the equation of a circle that is not in standard form:

\[ x^{2}+6 x+y^{2}+10 y+25=0\]

We need to restate this relationship so that the center and radius can be easily determined from the equation. In order to do this, we need to complete the square for both the \(x\) and the \(y\) variables. There are a variety of methods for completing the square I will demonstrate one of these below.

Take the original equation and move any term that doesn't have a variable to the other side:

\[ x^{2}+6 x+y^{2}+10 y=-25\]

Then open a space after the \(6 x\) and the \(10 y:\)

The idea is that we want trinomial expressions for both \(x\) and \(y\) that are perfect squares. If we look at squaring binomial expressions, we can see that there is a pattern:

\begin{array}{l} (x+3)^{2}=(x+3)(x+3)=x^{2}+6 x+9 \\ (x+4)^{2}=(x+4)(x+4)=x^{2}+8 x+16 \\ (x+5)^{2}=(x+5)(x+5)=x^{2}+10 x+25 \\ (x+6)^{2}=(x+6)(x+6)=x^{2}+12 x+36 \end{array}

We can see that if we have \(x^{2}+6 x,\) then this would correspond to \((x+3)(x+3)\) or \((x+3)^{2}\). But there's a problem: \((x+3)^{2}\) is not equal to \(x^{2}+6 x\). It's equal to \(x^{2}+6 x+9 .\) We can't just add a 9 to the \(x^{2}+6 x,\) but we can add a 9 to both sides.

\[ x^{2}+6 x+9+y^{2}+10 y=-25+9\]

Similarly, to complete the square on the \(y,\) we see that \(y^{2}+10 y\) corresponds to \((x+5)(x+5)\) or \((x+5)^{2}\). Here, we would need to add 25 to both sides to create a perfect square.

\[ \begin{aligned} x^{2}+6 x+9+y^{2}+10 y+25 &=-25+9+25 \\ x^{2}+6 x+9+y^{2}+10 y+25 &=9 \\ (x+3)^{2}+(y+5)^{2} &=9 \end{aligned} \]

So, the center of the circle is (-3,-5) and the radius is 3 Sometimes it is not so obvious what the values of \(h\) and \(k\) should be in completing the square. Consider the equation below: \[ x^{2}+20 x+y^{2}+30 y+15=0 \] If we look back at the examples for squaring binomials, we can see the pattern that relates the coefficient of the linear term to the values for \(h\) and \(k\) \[ \begin{array}{l} (x+3)^{2}=(x+3)(x+3)=x^{2}+6 x+9 \\ (x+4)^{2}=(x+4)(x+4)=x^{2}+8 x+16 \\ (x+5)^{2}=(x+5)(x+5)=x^{2}+10 x+25 \\ (x+6)^{2}=(x+6)(x+6)=x^{2}+12 x+36 \end{array} \] Notice that the coefficient of the linear term is always double the value of the numeral in the parentheses and the constant term is always that number squared.

\[ (x+n)^{2}=(x+n)(x+n)=x^{2}+2 n x+n^{2} \] So, given the problem above, \(x^{2}+20 x+y^{2}+30 y+15=0,\) first we can move the 15 to the right hand side of the equation and then complete the squares on the \(x\) and \(y\) variables. \[ \begin{array}{c} x^{2}+20 x+y^{2}+30 y=-15 \\ (x+?)^{2}+(y+?)^{2} \\ x^{2}+20 x+100+y^{2}+30 y+225=-15+100+225 \\ (x+10)^{2}+(y+15)^{2}=310 \end{array} \] We can now see that the center of the circle is at the point (-10,-15) and the radius of the circle is \(\sqrt{310} \approx 17.6\)

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