Math Vale

🏰 Geometry Kingdom · Geometry

Equations of Circles

Write the equation of a circle from its centre and radius, read the centre and radius back from an equation, complete the square to find them, and test whether a point lies on the circle.

In short

  • A circle is every point a fixed distance from a centre, so its equation is the distance formula squared: (x - h)2 + (y - k)2 = r2.
  • The brackets subtract, so (x + 6)2 means h = -6. Flip the sign every time you read a centre out of an equation, and every time you write one in.
  • The right-hand side is r2, never r and never the diameter. Halve a diameter before squaring, and take one square root at the end if the question wants r itself.
  • To undo general form, complete the square on the x terms and the y terms separately, adding the same amounts to the right-hand side as you go.

A circle is one sentence about distance

Keeper Orin measures the round tarn at Circle Basin with a cord pegged at the middle. Wherever the far end of the cord lands, that point is on the rim. That is the whole definition: a circle is every point at a fixed distance from a fixed centre.

Put that sentence on a grid and it turns into algebra. Let the centre be (h, k) and let (x, y) be any point on the rim. The horizontal gap between them is x - h and the vertical gap is y - k, and those two gaps are the legs of a right triangle whose hypotenuse is the cord itself. Pythagoras finishes the job:

(x - h)2 + (y - k)2 = r2

Nothing has been invented here. That is the distance formula with the distance already squared, and squaring is a kindness — it keeps every number whole and gets rid of the square-root sign for good.

When the centre is the origin the two gaps are just x and y, and the equation shrinks to its shortest form:

x2 + y2 = r2

So a circle of radius 3 about the origin is x2 + y2 = 9. Notice the right-hand side already: 9, not 3.

Reading and writing centre-radius form

Written as (x - h)2 + (y - k)2 = r2, the equation carries three facts, and each one has a trap.

The centre hides behind a subtraction. The form says *minus* h, so whatever sign you see inside the bracket, the coordinate is the opposite:

(x - 3)2 means h = 3 (x + 3)2 means x - (-3), so h = -3 x2 means h = 0

That is the single most common slip in the whole topic. Read (x + 6)2 + (y - 1)2 = 4 and the centre is (-6, 1) — not (6, -1), and not (6, 1).

The right-hand side is r squared. In x2 + y2 = 49 the radius is 7, because 49 is r2. Going the other way, a radius of 7 is written as 49. If a question hands you a *diameter*, halve it before you square: a diameter of 10 gives r = 5 and r2 = 25, never 100.

A radius can be given as a root. A circle with centre (2, -1) and radius sqrt(13) has equation (x - 2)2 + (y + 1)2 = 13, because squaring a square root simply undoes it. It is not 169, and the root does not stay on the page.

Completing the square: the disguised circle

Multiply the brackets of (x - 3)2 + (y + 2)2 = 25 out and tidy up and you get x2 + y2 - 6x + 4y - 12 = 0. Same circle, unrecognisable. This is called general form, and the way back is completing the square — once on the x terms, once on the y terms.

x2 + y2 - 6x + 4y - 12 = 0 (x2 - 6x) + (y2 + 4y) = 12 group, and move the number across half of -6 is -3, and (-3)2 = 9 half of 4 is 2, and 2 2 = 4 (x - 3)2 + (y + 2)2 = 12 + 9 + 4 add both to BOTH sides (x - 3)2 + (y + 2)2 = 25 centre (3, -2), r = 5

Two habits make this reliable. First, halve, then square: the halved number is what goes inside the bracket, and its square is what you add. Second, whatever you add on the left you add on the right — the right-hand side always grows, and forgetting that is what turns a radius of 5 into a radius of sqrt(12).

A shortcut worth knowing: from x2 + y2 + ax + by + c = 0 the centre is always (-a/2, -b/2). If a or b is odd the centre lands on a half, such as (1.5, -3), which is perfectly allowed — a centre does not have to sit on a grid crossing.

Testing a point, and meeting the axes

Is this point on the circle? Put its coordinates into the left-hand side and compare the result with the right-hand side. The left-hand side is a *squared* distance, so it must be compared with r2, never with r.

left = r2 -> on the circle left < r2 -> inside left > r2 -> outside

For P(7, -1) and (x - 3)2 + (y + 2)2 = 25: (7 - 3)2 + (-1 + 2)2 = 16 + 1 = 17, and 17 is less than 25, so P sits inside.

Where does it cross an axis? Every point on the x-axis has y = 0, so substitute y = 0 and solve what is left. For (x - 2)2 + (y - 4)2 = 25:

(x - 2)2 + (0 - 4)2 = 25 (x - 2)2 = 25 - 16 = 9 x - 2 = 3 or x - 2 = -3 x = 5 or x = -1

Two answers, because a square root has two signs. Sometimes there are none: the distance from the centre to the x-axis is simply the size of k, so compare |k| with r. Smaller than r and the circle cuts the axis twice; exactly r and it touches once (the axis is a tangent); larger than r and the circle never reaches it. That last idea also runs backwards — a circle centred at (5, -3) that touches the x-axis must have r = 3, so its equation is (x - 5)2 + (y + 3)2 = 9.

How the questions here ask you to answer

Circle Basin asks for four kinds of answer, and each has a fixed shape.

  • An equation is always multiple choice. A circle is not a function, so an equation cannot be typed and checked the way y = 2x + 1 can. Instead you pick it from a short list — and the wrong options are the four classic slips, so read all of them: the signs of h and k left as they appear, r written where r2 belongs, the diameter used as the radius, a point on the rim used as the centre.
  • A centre, a midpoint or a pair of intercepts is typed as a pair. Two boxes, "( , )". A centre is given as (h, k) in that order — the x bracket always supplies the first number, even when the equation prints the y bracket first. Intercepts are given as (x1, x2) with the smaller value first, and half answers such as (1.5, -3) are typed as decimals.
  • A radius is a whole number. Where r would be an untidy root, the question asks for r2 instead and expects a whole number: for (x - 5)2 + (y - 6)2 = 20 the answer to "what is r^2?" is 20.
  • On / inside / outside, and how a circle meets an axis, are multiple choice.

One more convention: r2 is a fine final answer, but r is a length, so it is never negative. If a question asks for the radius and you have r2 = 36, the answer is 6.

Worked examples

Example 1

Write the equation of the circle with centre (-4, 3) and diameter 10.

  1. The equation is built from the radius, so halve the diameter first: r = 10 / 2 = 5.
  2. The right-hand side is r2, so r2 = 25.
  3. The centre is subtracted, so h = -4 gives (x - (-4))2 = (x + 4)2, and k = 3 gives (y - 3)2.
  4. The equation is (x + 4)2 + (y - 3)2 = 25. (Choosing 100 would be squaring the diameter; choosing 5 would be forgetting to square at all.)

Example 2

A circle has centre C(-5, 2) and passes through P(-1, -1). Find r2, then write its equation.

  1. The radius is the distance from C to P, so r2 is the squared distance between them.
  2. The gaps are -1 - (-5) = 4 across and -1 - 2 = -3 up.
  3. r2 = 42 + (-3)2 = 16 + 9 = 25, so no square root is needed at all.
  4. Put the centre in the brackets with flipped signs: (x + 5)2 + (y - 2)2 = 25.

Example 3

Complete the square to find the centre and radius of x2 + y2 + 10x - 4y + 13 = 0.

  1. Group and move the number across: (x2 + 10x) + (y2 - 4y) = -13.
  2. Half of 10 is 5 and 52 = 25; half of -4 is -2 and (-2)2 = 4.
  3. Add 25 and 4 to both sides: (x + 5)2 + (y - 2)2 = -13 + 25 + 4 = 16.
  4. The centre is (-5, 2) — signs flipped out of the brackets — and r2 = 16, so r = 4.

Example 4

Where does the circle (x - 6)2 + (y - 8)2 = 100 cross the y-axis?

  1. Every point on the y-axis has x = 0, so substitute x = 0: (0 - 6)2 + (y - 8)2 = 100.
  2. 36 + (y - 8)2 = 100, so (y - 8)2 = 64.
  3. Take the square root of both sides, keeping both signs: y - 8 = 8 or y - 8 = -8.
  4. So y = 16 or y = 0, and the pair, smaller value first, is (0, 16).

Practice problems, with solutions

Three problems of increasing difficulty, each with the full working. In the game these are generated fresh every time; these three are fixed so this page always shows the same ones.

Problem 1

Difficulty 1 of 5

Which equation describes the circle with centre (1, 3) and radius 3?

  1. (x + 1)2 + (y + 3)2 = 9
  2. (x - 1)2 + (y - 3)2 = 9
  3. (x - 1)2 + (y - 3)2 = 3
  4. (x + 1)2 + (y + 3)2 = 3

Answer: B. (x - 1)2 + (y - 3)2 = 9

  1. Centre-radius form: (x - h)2 + (y - k)2 = r2.
  2. Here h = 1 and k = 3, so the brackets are (x - 1)2 and (y - 3)2.
  3. r = 3, so r2 = 9.
  4. The equation is (x - 1)2 + (y - 3)2 = 9.

Problem 2

Difficulty 3 of 5

A circle has equation (x - 6)2 + (y - 7)2 = 25. What is its radius?

Answer: 5

  1. The right-hand side is r2, so r2 = 25.
  2. r = sqrt(25) = 5.

Problem 3

Difficulty 4 of 5

The circle with centre C(-6, 5) passes through the point P(-7, 7). Which equation describes the circle?

  1. (x + 6)2 + (y - 5)2 = 5
  2. (x + 7)2 + (y - 7)2 = 5
  3. (x + 6)2 + (y - 5)2 = 3
  4. (x - 6)2 + (y + 5)2 = 5

Answer: A. (x + 6)2 + (y - 5)2 = 5

  1. r is the distance from the centre to any point on the circle, so r2 is the squared distance from C to P.
  2. r2 = (-7 - (-6))2 + (7 - 5)2 = (-1)2 + 22.
  3. r2 = 1 + 4 = 5.
  4. The equation is (x + 6)2 + (y - 5)2 = 5.

Common mistakes

  • Reading the centre of (x + 2)2 + (y - 7)2 = 9 as (2, -7). The signs inside the brackets are the opposite of the coordinates: the centre is (-2, 7).
  • Writing the radius where r2 belongs — an answer of ... = 5 for a radius of 5, when the right-hand side should read 25. The reverse also happens: saying the radius of x2 + y2 = 36 is 36 rather than 6.
  • Using a diameter as the radius. A diameter of 12 gives r = 6 and r2 = 36, not 144.
  • Completing the square on the left but leaving the right-hand side alone. The numbers you add to finish the brackets must be added to both sides, or the radius comes out too small.
  • Testing a point by comparing the squared distance with r instead of r2 — the left-hand side is already squared, so it belongs beside the right-hand side exactly as it is.

What you should be able to do

  • Write the equation of a circle from its centre and radius, or from its centre and a point on it.
  • Read the centre and radius from (x - h)2 + (y - k)2 = r2.
  • Complete the square to rewrite x2 + y2 + ax + by + c = 0 in centre-radius form.
  • Decide whether a point lies on, inside or outside a circle.

Where this fits in the curriculum

Common Core

  • HSG-GPE.A.1

    High school — Derive the equation of a circle of given centre and radius using the Pythagorean theorem; complete the square to find the centre and radius of a circle given by an equation.

  • HSG-GPE.B.4

    High school — Use coordinates to prove simple geometric theorems algebraically, for example that a given point lies on a given circle.

SAT

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