Center and Radius by Completing the Square

Fold the expanded form back into brackets.

An equation that hides the circle

The circle with center (2, −3) and radius 4 has the equation (x − 2)² + (y + 3)² = 16. Expand both brackets: x² − 4x + 4 + y² + 6y + 9 = 16. Collect the terms and bring everything to the left: x² + y² − 4x + 6y − 3 = 0.

That is the same circle, written multiplied out. In this form the center and the radius can no longer be read off: neither 2, −3 nor 4 appears anywhere in it. To find them, run the process backwards and put the brackets back. The tool for that is completing the square.

Gather the terms

Start from x² + y² − 4x + 6y − 3 = 0. Put the x terms next to each other and the y terms next to each other, and move the constant to the right by adding 3 to both sides: x² − 4x + y² + 6y = 3.

Complete the square twice

Complete the square on the x terms. Half of −4 is −2, and (x − 2)² = x² − 4x + 4. That is x² − 4x with an extra 4, so take the 4 away again: x² − 4x = (x − 2)² − 4.

Then do the same on the y terms. Half of 6 is 3, and (y + 3)² = y² + 6y + 9, so y² + 6y = (y + 3)² − 9. Each completed square leaves an extra constant to take away: here 4 and 9, the squares of the two halves.

x²2.5x2.5x(2.5)²x + 2.5

half of bx along each of two sides leaves a corner of side b/2 unfilled, which is why (b/2)² is added and then subtracted; the corner is 6.25

Make the missing corner 9

A square x² with a strip bx split in two, one half along each of two sides. The corner left unfilled is the constant that completes the square. Drag b to 6, the y² + 6y of the example with x in place of y, and the corner is 3² = 9.

Move the constants across

Put both completed squares into x² − 4x + y² + 6y = 3: (x − 2)² − 4 + (y + 3)² − 9 = 3. Add 4 and 9 to both sides: (x − 2)² + (y + 3)² = 3 + 4 + 9 = 16.

This is the center-radius form, (x − a)² + (y − b)² = r², again.

Read the center and the radius

In (x − 2)² + (y + 3)² = 16 the brackets are zero at x = 2 and y = −3, so the center is (2, −3). The right side is r² = 16, so the radius is √16 = 4.

Check with a point on the circle. The point 4 above the center is (2, 1), and (2 − 2)² + (1 + 3)² = 0 + 16 = 16. In the expanded equation it gives 4 + 1 − 8 + 6 − 3 = 0, as it should.

xy(2, −3)

x² + y² − 4x + 6y − 3 = 0 is the circle with center (2, −3) and radius 4. It passes through (2, 1), 4 above the center.

A second example

Find the center and radius of x² + y² − 4x + 2y − 20 = 0. Gather the terms and move the constant: x² − 4x + y² + 2y = 20.

Complete each square. Half of −4 is −2, so x² − 4x = (x − 2)² − 4. Half of 2 is 1, so y² + 2y = (y + 1)² − 1. Then (x − 2)² − 4 + (y + 1)² − 1 = 20, and adding 4 and 1 to both sides gives (x − 2)² + (y + 1)² = 25.

The center is (2, −1) and the radius is √25 = 5. In both examples the center is half of each coefficient of x and of y, with its sign changed: half of −4 is −2, which gives 2, and half of 2 is 1, which gives −1. That is a useful check on the answer.

When there is no circle

The number on the right after completing the squares is r², so it must be positive. If x² + y² − 4x + 6y + 13 = 0, the same steps give (x − 2)² + (y + 3)² = −13 + 4 + 9 = 0, and the only point that fits is (2, −3) itself. With + 20 in place of + 13 the right side would be −7, and no point fits at all, because two squares cannot add up to a negative number.

Completing the square in this way needs x² and y² to have a coefficient of 1. If the equation starts 2x² + 2y², divide every term by 2 first.

The usual mistakes

Getting the sign of the center wrong. (x − 2)² + (y + 3)² = 16 has its center at (2, −3), not (−2, 3): the center is the point that makes both brackets zero.

Not halving. The center of x² + y² − 4x + 6y − 3 = 0 is not (4, −6). Completing the square halves each coefficient first, so the brackets are (x − 2) and (y + 3).

Adding the extra constant instead of taking it away. (x − 2)² already contains the 4, so x² − 4x = (x − 2)² − 4, not (x − 2)² + 4.

Forgetting to move the extra constants across. Stopping at (x − 2)² + (y + 3)² = 3 would give a radius of √3. Both 4 and 9 must be added to the right side, which makes it 16.

Taking 16 as the radius. It is r², so the radius is √16 = 4.

Worked example: The Center and Radius of a Roundabout from Its Expanded Equation

Question On the plan of a road junction, distances are in meters, with x measured to the east and y to the north. The curb of a circular roundabout has the equation x2 + y2 − 12x − 16y + 84 = 0. (a) Find the center and the radius of the roundabout. (b) A lamp post stands at the origin. How far is the lamp post from the nearest point of the curb?

  1. 1.Group the x terms and the y terms, and subtract 84 from both sides: (x2 − 12x) + (y2 − 16y) = −84.

    0246810121402468101214meters east, xmeters north, ylamp postx2+ y2− 12x − 16y + 84 = 0(x2− 12x) + (y2− 16y) = −84
    0246810121402468101214meters east, xmeters north, ylamp postx2+ y2− 12x − 16y + 84 = 0(x2− 12x) + (y2− 16y) = −84
    Group the x terms and the y terms, and take the constant to the other side: (x2 − 12x) + (y2 − 16y) = −84.

  2. 2.Complete each square. Half of 12 is 6, and (x − 6)2 = x2 − 12x + 36, so x2 − 12x = (x − 6)2 − 36. Half of 16 is 8, so y2 − 16y = (y − 8)2 − 64.

    0246810121402468101214meters east, xmeters north, ylamp postx2− 12x = (x − 6)2− 36y2− 16y = (y − 8)2− 64
    0246810121402468101214meters east, xmeters north, ylamp postx2− 12x = (x − 6)2− 36y2− 16y = (y − 8)2− 64
    Complete each square: x2 − 12x = (x − 6)2 − 36 and y2 − 16y = (y − 8)2 − 64.

  3. 3.Substitute both: (x − 6)2 − 36 + (y − 8)2 − 64 = −84. Add 36 and 64 to both sides: (x − 6)2 + (y − 8)2 = 16.

    0246810121402468101214meters east, xmeters north, ylamp post(x − 6)2+ (y − 8)2= −84 + 36 + 64(x − 6)2+ (y − 8)2= 16 = 42
    0246810121402468101214meters east, xmeters north, ylamp post(x − 6)2+ (y − 8)2= −84 + 36 + 64(x − 6)2+ (y − 8)2= 16 = 42
    Substitute and add 36 and 64 to both sides: (x − 6)2 + (y − 8)2 = 16.

  4. 4.(a) Since 16 = 42, the roundabout has its center at (6, 8) and a radius of 4 m.

    0246810121402468101214meters east, xmeters north, y4 mcenter (6, 8)lamp post(x − 6)2+ (y − 8)2= 42the center is (6, 8) and the radius is 4 m
    0246810121402468101214meters east, xmeters north, y4 mcenter (6, 8)lamp post(x − 6)2+ (y − 8)2= 42the center is (6, 8) and the radius is 4 m
    (a) This is a circle with center (6, 8) and radius √16 = 4 m.

  5. 5.The lamp post at the origin is √62 + 82 = √100 = 10 m from the center. This is more than the radius, so the lamp post is outside the roundabout, and the nearest point of the curb lies on the straight line from the lamp post to the center.

    0246810121402468101214meters east, xmeters north, y4 m(6, 8)lamp postlamp post to center: 62+ 82= 100, and√100 = 10 mthe nearest point of the curb is on this line
    0246810121402468101214meters east, xmeters north, y4 m(6, 8)lamp postlamp post to center: 62+ 82= 100, and√100 = 10 mthe nearest point of the curb is on this line
    The lamp post at the origin is √62 + 82 = 10 m from the center. The nearest point of the curb lies on the line from the lamp post to the center.

  6. 6.(b) Along that line the curb is one radius nearer than the center, so the nearest point of the curb is 10 − 4 = 6 m from the lamp post.

    0246810121402468101214meters east, xmeters north, y4 m(6, 8)6 mlamp post10 − 4 = 6 mthe curb is 6 m from the lamp post at its nearest
    0246810121402468101214meters east, xmeters north, y4 m(6, 8)6 mlamp post10 − 4 = 6 mthe curb is 6 m from the lamp post at its nearest
    (b) The nearest point of the curb is one radius short of the center: 10 − 4 = 6 m from the lamp post.

Answer: (a) Center (6, 8) and radius 4 m; (b) 6 m

Common mistakes

  • Reading the center as (−6, −8) or as (12, 16). In (x − 6)2 + (y − 8)2 = 16 the center is the point that makes both brackets zero, which is (6, 8): half of each coefficient, with the sign changed.
  • Giving the radius as 16. The right-hand side of the equation is r2, so the radius is √16 = 4 m.

More graphs of equations problems, worked step by step →

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