EQUATIONS OF CIRCLES

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While the important technique of completing the square is on your fingertips,
it's a good time to discuss the equations of circles.

Suppose that a circle has center (h,k) and radius r .
Let (x,y) be a typical point on this circle.
Then, the distance from (x,y) to (h,k) must equal r :

circle with center (h,k) and radius r

(x -h)2 + (y-k )2 =r	(recall from The Distance Formula that the distance between points (x1, y1) and (x2, y2) is given by the formula (x2 -x1) 2 + (y2- y1) 2 )
(x-h )2 + (y-k) 2=r 2	(square both sides)

Thus, we have:

EQUATIONS OF CIRCLES (standard form)

The equation of the circle with center (h,k) and radius r is:

(x-h )2 + (y-k) 2=r 2

For example, (x-3 )2 + (y+5) 2=15 is the circle with center (3,-5) and radius 15 .
Students sometimes like to think of it this way:

the x-value of the center of the circle is the number that makes (x-3) equal to zero (ANSWER: 3)
the y-value of the center of the circle is the number that makes (y+5) equal to zero (ANSWER: -5)
the radius is the square root of the right-hand side

Unfortunately, circles aren't always given to you in this nice form.
Sometimes they're given to you all multiplied out.
In this situation, you need to be able to recognize that you're dealing with a circle,
and use the technique of completing the square to get it into standard form.

Multiplying out the left-hand side of (x-h )2 + (y-k) 2=r 2 gives:

x2 -2hx+ h2+ y2- 2ky+k 2=r 2
The key observation is that there are only five types of terms: x2 , y2 , x, y, and constant.
Also, when the x2 and y2 terms are on the same side of the equation, they have the same coefficient.
These observations lead to the following way to recognize circles:

RECOGNIZING CIRCLES

Let b , c , and d be real numbers, and let a be a nonzero real number.
Equations of the form
ax2 +ay2 +bx+cy =d graph as circles.

You must have both x2 and y2 terms,
and they must have the same coefficient, when they are on the same side of the equation.
You are allowed (but not required) to have x , y , and constant terms.

It is possible to end up with a circle with radius zero,
sometimes called a "point" circle, like x2 +y2 =0 .
The only solution to this equation is (0,0) .

It is also possible to end up with an "imaginary" circle, like x2 +y2 =-1 .
Notice that there are no real numbers x and y that make this equation true.

The following example illustrates the process of recognizing the equation of a circle
and putting it in standard form so that it can be easily graphed:

3x 2-5x -7=1-3 y2	(original equation: squared terms will have same coefficient when on same side; allowed (but not required) to have x , y , and constant terms; there are no other types of terms)
3x 2-5x+ 3y2 =8	(put all variable terms on the left, and constant terms on the right)
x2 -53 x+y 2=83	(divide both sides by 3; coefficient of squared term must be 1 to use the technique of completing the square)
x2 -53 x+ ( -52⋅3 )2 +y2 =8 3+25 36	(on both sides, add the appropriate number to complete the square)
(x-5 6) 2+y 2= 12136	(rename the perfect square trinomial; add fractions)

Thus, this is the circle with center (5 6,0) and radius 121 36= 116 .

For confidence in your work, use your calculator to check at least one of the four easy points on the circle in the original equation—say, the point (5 6,11 6) :

3x 2-5x -7=1-3 y2

3(5 6) 2-5( 56) -7= ?1-3( 116 )2

-109 12=-109 12

Yes!!

On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.

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