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How do we write the equation of a circle and find its centre and radius?

Write the equation of a circle given the centre and radius, and find the centre and radius from a general equation by completing the square

A focused answer to the N(A)-Level Additional Mathematics outcome on circles. The standard form of a circle equation, finding the centre and radius, and converting from the general expanded form by completing the square.

Generated by Claude Opus 4.88 min answer

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
  4. Try this

What this dot point is asking

SEAB wants you to work with the equation of a circle: write it down given the centre and radius, and go the other way, finding the centre and radius from an expanded "general" equation by completing the square. This ties the circle work directly to the completing-the-square skill from the algebra strand, and it lets you answer questions about points, lines and circles together.

The answer

The standard form

A circle with centre (a,b)(a, b) and radius rr has the equation:

(x−a)2+(y−b)2=r2(x - a)^2 + (y - b)^2 = r^2

Every point (x,y)(x, y) on the circle is a distance rr from the centre, which is exactly what the distance formula written this way says. Watch the signs: a centre of (2,−3)(2, -3) gives (x−2)2+(y+3)2(x - 2)^2 + (y + 3)^2.

Writing the equation from centre and radius

Substitute the centre for (a,b)(a, b) and the radius for rr, remembering to square the radius on the right. Centre (2,−3)(2, -3), radius 55 gives (x−2)2+(y+3)2=25(x - 2)^2 + (y + 3)^2 = 25.

The general (expanded) form

Multiplying out the standard form gives a general equation of the type:

x2+y2+2gx+2fy+c=0x^2 + y^2 + 2gx + 2fy + c = 0

You can recognise a circle because the x2x^2 and y2y^2 coefficients are equal (both 11 here) and there is no xyxy term.

Finding the centre and radius from the general form

To recover the centre and radius, complete the square separately on the xx terms and the yy terms:

  1. Group: (x2+2gx)+(y2+2fy)=−c(x^2 + 2gx) + (y^2 + 2fy) = -c.
  2. Complete each square, subtracting the squares of the halves.
  3. Move the subtracted constants to the right-hand side.
  4. Read off the centre (a,b)(a, b) and radius rr from the resulting standard form.

Examples in context

Example 1. Does a point lie on, inside or outside? Substitute a point into the left-hand side (x−a)2+(y−b)2(x - a)^2 + (y - b)^2 and compare with r2r^2: equal means on the circle, smaller means inside, larger means outside. The standard form turns a position question into one substitution.

Example 2. Tangent meets circle once. A line is a tangent to a circle when it meets the circle at exactly one point. Substituting the line into the circle equation gives a quadratic, and the tangency condition is that its discriminant is zero, linking circles to the discriminant from the quadratics strand.

Try this

Q1. Write the equation of the circle with centre (0,0)(0, 0) and radius 44. [1 mark]

  • Cue. x2+y2=16x^2 + y^2 = 16.

Q2. State the centre and radius of (x−1)2+(y+5)2=9(x - 1)^2 + (y + 5)^2 = 9. [2 marks]

  • Cue. Centre (1,−5)(1, -5), radius 33.

Q3. Find the centre and radius of x2+y2−4x−6y+9=0x^2 + y^2 - 4x - 6y + 9 = 0. [3 marks]

  • Cue. (x−2)2+(y−3)2=4(x - 2)^2 + (y - 3)^2 = 4; centre (2,3)(2, 3), radius 22.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original2 marksWrite down the equation of the circle with centre (2,−3)(2, -3) and radius 55.
Show worked answer →

The standard form is (x−a)2+(y−b)2=r2(x - a)^2 + (y - b)^2 = r^2 with centre (a,b)(a, b) and radius rr.

Here (a,b)=(2,−3)(a, b) = (2, -3) and r=5r = 5, so (x−2)2+(y+3)2=25(x - 2)^2 + (y + 3)^2 = 25.

What markers reward: using the standard form, handling the signs of the centre correctly (y−(−3)=y+3y - (-3) = y + 3), and squaring the radius to 2525.

Original4 marksFind the centre and radius of the circle x2+y2−6x+4y−12=0x^2 + y^2 - 6x + 4y - 12 = 0.
Show worked answer →

Group and complete the square: (x2−6x)+(y2+4y)=12(x^2 - 6x) + (y^2 + 4y) = 12.

(x−3)2−9+(y+2)2−4=12(x - 3)^2 - 9 + (y + 2)^2 - 4 = 12, so (x−3)2+(y+2)2=25(x - 3)^2 + (y + 2)^2 = 25.

Centre (3,−2)(3, -2), radius 25=5\sqrt{25} = 5.

What markers reward: completing the square on both xx and yy, moving the subtracted constants across correctly, and reading the centre and radius from the standard form.

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