Circles: Working with the Equations
Purplemath
The useful form of the circle equation, in my opinion, is the center-radius form; namely, (x − h)2 + (y − k)2 = r2.
This is the useful form because you can read the coordinates of the center and the length of the radius directly from the equation; the center is at the point (h, k) and the radius as a length of r units.
Once you have become familiar with this equation, you'll be asked to find information from the equation, or else to create the equation from given information, among other things.
- State the radius and center of the circle with equation 16 = (x − 2)2 + (y − 3)2.
The numerical side, the 16, is the *square* of the radius, so it actually indicates 16 = r2 = 42, so the radius is r = 4.
The first of the two squared-variable parts is (x − 2)2. When finding the coordinates of the center, I only need to look at the insides, so x − 2 is the part that I care about. The circle-radius form of the circle equation says that the x-coordinate of the center is h. In this case, that means that the coordinate is 2.
By the same reasoning, the second of the two squared-variable parts tells me that the y-coordinate of the center is h = 3.
center: (h, k) = (2, 3)
radius: r = 4
Keep in mind that the center-radius form of the circle equation is:
(x − h)2 + (y − k)2 = r2
This means that, ifthere is a "plus" sign in the middle of one of the squared terms, they've actually subtracted a negative. So don't just mindlessly copy over the number; make sure you've got the correct sign.
- 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.
At this point, 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 subtraction in it, so I need to convert first to that form.
y + 3 = y − (−3)
So the y-coordinate of the center is actually k = −3.
radius: r = 5
center: (h, k) = (0, −3)
Be careful when you're reading the coordinates of the center from the equation. It's very easy to get in the habit of just copying down the numbers, while ignoring the signs in front of them. But those signs matter! If there is a "plus" sign in the middle of a squared-variable expression, remember that this means that the coordinate is actually a negative number.
In the previous examples, center and radius information was extracted from a given equation. You'll also need to be able to work from given information backwards to find an equation, and possibly also the center and the radius.
- Find an equation for the circle with center (h, k) = (4, −2) and radius r = 10.
They've given me the center and the radius. I have memorized the center-radius form of the circle equation. So I'll just plug the center and radius into the center-radius form:
(x − (4))2 + (y − (−2))2 = 102
I could stop with the above equation, but I'd like to square out the numerical side. (This is just a personal preference.) This gives me an answer of:
(x − 4)2 + (y + 2)2 = 100
Since no particular form of the equation was specified in the above exercise, the center-radius form of the circle equation is an acceptable answer. But if your textbook or instructor specifies some other particular format, then you may need to multiply things out completely. Multiplying and simplifying my answer from above looks like this:
x2 − 8x + 16 + y2 + 4y + 4 = 100
x2 + y2 − 8x + 4y + 20 = 100
x2 + y2 − 8x + 4y − 80 = 0
Keep in mind that there is no standard meaning to the term "standard form". If your book specifies something as being "standard" form, then memorize that form for your tests. And don't be surprised if a different class uses as different "standard" form.
- Find the center and radius of the circle with equation x2 + y2 + 2x + 8y + 8 = 0
To convert to center-radius form, I'll need to complete the squares. (If you don't remember how to complete the square, then review now.)
x2 + y2 + 2x + 8y + 8 = 0
x2 + 2x + [ ] + y2 + 8y + [ ] = −8 + [ ] + [ ]
x2 + 2x + [ 1 ] + y2 + 8y + [16] = −8 + [ 1 ] + [16]
(x + 1)2 + (y + 4)2 = 9
(x − (−1))2 + (y − (−4))2 = 32
I made a point, in that last line, of converting the addition inside the two squared-variable expressions to subtraction of a negative. This is because (a) subtraction is the format of the circle-radius form and (b) it helps me remember the "minus" signs that I need for my answer to be correct.
center: (h, k) = (−1, −4)
radius: r = 3
- Find an equation for the circle centered at (−5, 12) and passing through the point (−2, 8).
The Distance Formula, measuring between the center and any point on the circle, will return the value of the radius; this is because the radius is, by definition, the distance between the center and any point on the circle.
So I'll plug the given values into the Distance Formula, and simplify:
![d = sqrt[(−5 − (−2))^2 + (12 − 8)^2] = sqrt[(-3)^2 + (4)^2] = sqrt[9 + 16] = sqrt[25] = 5 = r](conics/circle05.gif)
Then the center-radius form of the equation is:
(x − (−5))2 + (y − (12))2 = 52
They didn't ask me for the center (indeed, they gave me the center in the original exercise statement), so I don't need the equation to be in "subtraction of a negative" form. And they didn't ask me for the length of the radius, so I don't need the squared form of the number, either. So I could leave my answer as the above, but I'll simplify inside the parentheses, just so it looks a little nicer. (Again, this is just a personal preference.)
(x + 5)2 + (y − 12)2 = 25
Warning: It is very easy to forget that sign in the middle of the squared parts. Don't be careless: Those two squared expressions are added together, so make sure that the appropriate sign is between them!
- Find an equation for the circle with a diameter with endpoints at (9, −4) and (−1, 0).
While it might not seem like it, at first glance, they *have* given me enough information. A center's diameter is a line from one side of the circle to the other, passing through the center. The radius is half the length of the diameter, and the center is at the diameter's midpoint. The Midpoint Formula gives me:
I can use either end of the diameter for my point on the circle; the distance between the center and the circle will be the same, regardless. I like smaller numbers, so I'll pick (−1, 0). The Distance Formula gives me:
![d = sqrt[(4+1)^2 + (-2-0)^2] = sqrt[(5)^2 + (-2)^2] = sqrt[25 + 4] = sqrt[29]](conics/circle07.gif)
This distance is the length of the radius r, so r2 = 29. Plugging my results into the center-radius form, I get:
(x − (4))2 + (y − (−2))2 = 29
I'll simplify inside the parentheticals to get my final answer:
(x − 4)2 + (y + 2)2 = 29
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