CIR - Equations of Circles Lesson

Equations of Circles

You have been seeing circles your whole life, and have even studied them in previous courses. However, we are going to go further with the study of circles - what do they look like in the coordinate plane? What is the equation of a circle? Where did it come from? Let's start there... what is the equation of a circle and where did it come from.

The standard equation of a circle is  $\left(x-h\right)^2+\left(y-k\right)^2=r^2$, where (h, k) is the center of the circle and r  is the radius.

For example, a circle with center (3, 2) and radius 5 would have the following equation: 

$\left(x-3\right)^2+\left(y-2\right)^2=5^2$

$\left(x-3\right)^2+\left(y-2\right)^2=25$ 

What would be the equation of a circle with center (-4, 1) and radius 3?

Solution:

$\left(x+4\right)^2+\left(y-1\right)^2=3^2$

$\left(x+4\right)^2+\left(y-1\right)^2=9$

But where does this equation come from?

Consider the circle to the right. We have the center labeled as (h, k), the radius labeled "r," and we have a point on the circle labeled (x, y). We call the shorter leg of the right triangle formed "a," and the longer leg "b".

By the Pythagorean Theorem, we know  $a^2+b^2=r^2$.

The length of "a" is the distance from the x-coordinate of the point (x, y) to the x-coordinate of the center (h, k). This is represented by |x - h|. Likewise, the length of "b" is the distance from the y-coordinate of the point (x, y) to the y-coordinate of the center (h, k). This is represented by |y - k|. Plug these in for a and b in the original formula, and you'll see the equation of a circle: $\left(x-h\right)^2+\left(y-k\right)^2=r^2$

Notice that we dropped the absolute value signs when we plugged them into the equation. Why do you think this is? 

• Solution: The absolute value signs were there initially because distance is always positive we can't have a negative distance. However, since we are squaring these values in the equation, they will automatically become positive anyways, so the absolute value signs are no longer needed!

Try the practice problems below to see how much you know!

1. What is the equation of a circle with center (-2, -1) and radius 5?

• Solution: $\left(x+2\right)^2+\left(y+1\right)^2=25$

2. What is the equation of a circle with center (4, -6) and radius 1? 

• Solution: $\left(x-4\right)^2+\left(y+6\right)^2=1$

3. What is the equation of this circle? 

• Solution: 
  • center = (0,0)
  • radius = 2
  • $x^2+y^2=4$

4. Name the center and radius of the circle with equation $\left(x-2\right)^2+\left(y+1\right)^2=36$  . 

• Solution: 
  • center = (2, -1)
  • radius = 6

5. What is the equation of this circle? 

• Solution: 
  • center = (0, 3)
  • radius = 4
  • $x^2+\left(y-3\right)^2=16$

Unfortunately, circle equations are not always in this nice, easy-to-use standard form. Sometimes, the equation is given to you when it is all multiplied out in an expanded form, called the general form. This form can be created by expanding out the standard form.  When we do so, we get an equation of the form $ax^2+by^2+cx+dy+e=0$. Since we cannot find the center and radius directly from this form, we can convert it back to standard form by completing the square! Before we get started, watch the video below for a refresher on completing the square.

Let's try an example.

Example

Let's convert the circle equation $x^2+y^2-2x-4y-4=0$ to standard form to find the center and radius. To do this, we must complete the square.

Now it's time to test yourself! Before checking the answers on each step, try it on your own to see how you do!

Find the center and radius of the circle defined by the following equation:

 $4x^2+4y^2-16x-24y+51=0$

Step 1 : Be sure the coefficients of both the x²  and y²  terms are 1. If not, divide the whole equation by whatever the coefficient is. After trying this step on your own, check your answer with the solution shown below.

• Solution: Since $x^2$ and $y^2$ terms both have coefficient of 4, we must divide the whole equation by 4 to get:
  • $x^2+y^2-4x-6y+\frac{51}{4}=0$

Step 2 : Gather the x's together and the y's together and move the constant to the right side of the equation. After trying this step on your own, check your answer with the solution shown below.

• Solution:  $\left(x^2-4x\right)+\left(y^2-6y\right)=-\frac{51}{4}$

Step 3 : Complete the square for the x's and for the y's. Remember, this means that for both the x's and the y's, you need to add $\left(\frac{b}{2}\right)^2$ inside the parentheses, as well as to the other side of the equation to cancel it out. After trying this step on your own, check your answer with the solution shown below.

• Solution:  $\left(x^2-4x+4\right)+\left(y^2-6y+9\right)=-\frac{51}{4}+4+9$

• • for the x's, $\left(\frac{b}{2}\right)^2=\left(\frac{-4}{2}\right)^2=\left(-2\right)^2=4$
  • for the y's, $\left(\frac{b}{2}\right)^2=\left(\frac{-6}{2}\right)^2=\left(-3\right)^2=9$

Then you factor the "x" expression and the "y" expression and simplify the right side of the equation. Remember, each expression is now a difference of squares, and they factor to $\left(x+\frac{b}{2}\right)^2\:and\:\left(y+\frac{\:b}{2}\right)^2$ . After trying this step on your own, check your answer with the solution shown below.

• Solution: $\left(x-2\right)^2+\left(y-3\right)^2=\frac{1}{4}$

Now that our circle equation is in standard form, we can find the center and radius! After trying this step on your own, check your answer with the solution shown below.

• Solution: $\left(x-2\right)^2+\left(y-3\right)^2=\frac{1}{4}$
  • $r^2=\frac{1}{4}\longrightarrow radius=\frac{1}{2};\:center=\left(2,\:3\right)$

The last thing we want to look at with circles and their equations is how to tell if a given point lies on a particular circle.

For example, does the point (2, -1) lie on the circle with center (2, -5) and radius 6? Let's check...

Remember the equation of a circle is $\left(x-h\right)^2+\left(y-k\right)^2=r^2$.

Now plug in the given center and radius. $\left(x-2\right)^2+\left(y+5\right)^2=6^2$.

Now, plug in the given point for x and y to see if we get a true statement. If we do, the point lies on the circle. If not, it doesn't.

$\left(2-2\right)^2+\left(-1+5\right)^2=6^2$

$0^2+4^2=6^2$

$16<36$

This is false. Therefore the point (2, -1) lies inside the circle since 16 < 36!

Let's try one more.

Does the point (0, -8) lie on the circle $x^2+y^2=64$ ?

This one is even simpler, since the equation is given to us. Just plug in the point for x and y into the equation and see if you get a true statement:

$0^2+\left(-8\right)^2=64$

$0+64=64$

$64=64$

So the point (0, -8) does lie on the circle.

What about the point (2, -7)? Does it lie on the circle $x^2+y^2=64$ ?

Let's check.

Plug in 2 for the x and -7 for the y.

$2^2+\left(-7\right)^2=64$

$4+49=64$

$53\ne64$

So (2, -7) does NOT lie on the circle.

Practice your knowledge of this concept below:

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