Equation of a Circle

This lesson covers the equation of a circle in the coordinate plane. Circles appear frequently in A-Level Mathematics, both as standalone problems and as part of larger questions involving tangents, intersections, and parametric curves. You must be able to work with both forms of the equation and convert between them fluently.

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Standard Form of the Equation of a Circle

A circle with centre $(a, b)$(a,b) and radius $r$r has equation

(x − a)² + (y − b)² = r²

This follows directly from the distance formula: a point (x, y) lies on the circle if and only if its distance from the centre (a, b) is exactly r.

Special case: If the centre is at the origin (0, 0), the equation simplifies to

x² + y² = r²

Example 1: Write down the equation of the circle with centre (3, −2) and radius 5.

(x − 3)² + (y + 2)² = 25

Example 2: State the centre and radius of the circle (x + 4)² + (y − 1)² = 49.

Centre = (−4, 1)    (note the signs carefully)
Radius = √49 = 7

Exam Tip: Be very careful with signs. If the equation contains (x + 4)², the x-coordinate of the centre is −4, not 4. Write out the comparison with the standard form explicitly.

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Expanded Form

If we expand the standard form, we obtain the general or expanded form:

x² + y² + 2gx + 2fy + c = 0

where the centre is $(-g, -f)$(−g,−f) and the radius is $\sqrt{g^2 + f^2 - c}$g2+f2−c​.

This form arises when the brackets are expanded. For the equation to represent a circle, we need $g^2 + f^2 - c > 0$g2+f2−c>0.

Example 3: Expand (x − 3)² + (y + 2)² = 25.

x² − 6x + 9 + y² + 4y + 4 = 25
x² + y² − 6x + 4y + 13 = 25
x² + y² − 6x + 4y − 12 = 0

Here g = −3, f = 2, c = −12. Centre = (3, −2), radius = √(9 + 4 + 12) = √25 = 5. ✓

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Completing the Square to Find Centre and Radius

When given the expanded form, you must complete the square on both x and y terms to recover the centre and radius.

Method:

1. Group the x terms and y terms together.
2. Complete the square on each group.
3. Rearrange into standard form.

Example 4: Find the centre and radius of the circle x² + y² − 8x + 6y = 0.

Step 1: Group the terms.
(x² − 8x) + (y² + 6y) = 0

Step 2: Complete the square.
(x² − 8x + 16) − 16 + (y² + 6y + 9) − 9 = 0
(x − 4)² + (y + 3)² = 25

Step 3: Read off centre and radius.
Centre = (4, −3), radius = 5

Example 5: Find the centre and radius of x² + y² + 10x − 2y + 17 = 0.

(x² + 10x) + (y² − 2y) = −17
(x² + 10x + 25) − 25 + (y² − 2y + 1) − 1 = −17
(x + 5)² + (y − 1)² = −17 + 25 + 1 = 9

Centre = (−5, 1), radius = 3

Example 6: Show that x² + y² − 4x + 6y + 15 = 0 does not represent a circle.

(x² − 4x + 4) + (y² + 6y + 9) = −15 + 4 + 9 = −2
(x − 2)² + (y + 3)² = −2

Since the right-hand side is negative, this does not represent a circle (no real points satisfy this equation).

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Determining Whether a Point Lies On, Inside, or Outside a Circle

Substitute the point into the left-hand side of the equation:

• If $(x - a)^2 + (y - b)^2 = r^2$(x−a)2+(y−b)2=r2, the point lies on the circle.
• If $(x - a)^2 + (y - b)^2 < r^2$(x−a)2+(y−b)2<r2, the point lies inside the circle.
• If $(x - a)^2 + (y - b)^2 > r^2$(x−a)2+(y−b)2>r2, the point lies outside the circle.

Equivalently, compare the distance from the point to the centre with the radius.

Example 7: Determine whether P(1, 2) lies on, inside, or outside the circle (x − 3)² + (y + 1)² = 16.

(1 − 3)² + (2 + 1)² = 4 + 9 = 13
Since 13 < 16, the point lies inside the circle.

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Finding the Equation of a Circle Given Conditions

Given three points on the circle

Substitute each point into x² + y² + 2gx + 2fy + c = 0 to obtain three simultaneous equations in g, f, and c.

Example 8: Find the equation of the circle passing through A(0, 0), B(6, 0), and C(0, 8).

Substituting into x² + y² + 2gx + 2fy + c = 0:

A(0, 0): 0 + 0 + 0 + 0 + c = 0  ⟹  c = 0
B(6, 0): 36 + 0 + 12g + 0 + 0 = 0  ⟹  g = −3
C(0, 8): 0 + 64 + 0 + 16f + 0 = 0  ⟹  f = −4

Equation: x² + y² − 6x − 8y = 0
Centre = (3, 4), radius = √(9 + 16) = 5

Given the endpoints of a diameter

If A and B are endpoints of a diameter, then the centre is the midpoint of AB and the radius is half the length of AB.

Example 9: A circle has diameter from P(−1, 3) to Q(5, −1). Find its equation.

Centre = ((−1+5)/2, (3+(−1))/2) = (2, 1)
Diameter = √[(5−(−1))² + (−1−3)²] = √[36 + 16] = √52
Radius = √52 / 2 = √13

Equation: (x − 2)² + (y − 1)² = 13

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Worked Examination-Style Problem

Problem: The circle C has equation x² + y² − 6x + 2y − 15 = 0.

(a) Find the centre and radius of C.

(b) Show that the point A(7, 2) lies on C.

(c) Find the equation of the tangent to C at A.

Solution:

(a)

(x² − 6x) + (y² + 2y) = 15
(x − 3)² − 9 + (y + 1)² − 1 = 15
(x − 3)² + (y + 1)² = 25

Centre = (3, −1), radius = 5

(b)

(7 − 3)² + (2 + 1)² = 16 + 9 = 25 = r²  ✓
So A(7, 2) lies on C.

(c) The tangent at A is perpendicular to the radius CA.

Gradient of CA = (2 − (−1))/(7 − 3) = 3/4
Gradient of tangent = −4/3

Tangent: y − 2 = −4/3 (x − 7)
3(y − 2) = −4(x − 7)
3y − 6 = −4x + 28
4x + 3y − 34 = 0

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Summary

• Standard form: (x − a)² + (y − b)² = r², centre (a, b), radius r.
• Expanded form: x² + y² + 2gx + 2fy + c = 0, centre (−g, −f), radius √(g² + f² − c).
• Complete the square to convert from expanded form to standard form.
• To check if a point is on/inside/outside, substitute and compare with r².
• To find the equation given three points, substitute into the general form and solve simultaneously.
• Diameter endpoints: centre is the midpoint, radius is half the diameter length.

Exam Tip: When completing the square, lay out your working carefully — do the x terms first, then the y terms, and collect the constants. Always verify your answer by checking the radius-squared is positive. If the question asks you to "find the centre and radius", state them explicitly as a coordinate pair and a value.

A-Level Deep Dive: Equation of a Circle

Spec mapping

AQA 7357 Paper 1 — Pure Mathematics, Section C: Coordinate Geometry. The specification requires students to "use the equation of a circle in the form $(x-a)^2 + (y-b)^2 = r^2$(x−a)2+(y−b)2=r2" and "complete the square to find the centre and radius of a circle from the general form $x^2 + y^2 + 2gx + 2fy + c = 0$x2+y2+2gx+2fy+c=0." Beyond Section C, circle equations recur in Section B (Algebra and functions, completing the square), Section E (Trigonometry, parametric form $x = a + r\cos\theta$x=a+rcosθ, $y = b + r\sin\theta$y=b+rsinθ), Section J (Differentiation, implicit differentiation of $x^2 + y^2 = r^2$x2+y2=r2) and Section L (Vectors, the magnitude condition $|\mathbf{r} - \mathbf{c}| = r$∣r−c∣=r). The AQA formula booklet provides the standard form but expects fluency with the general form and tangent-perpendicular-to-radius condition without prompting.

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Worked example with full mark scheme

Question (8 marks):

The circle $C$C has equation $x^2 + y^2 - 6x + 4y - 12 = 0$x2+y2−6x+4y−12=0.

(a) Find the centre and radius of $C$C. (3)

(b) The point $P(7, 3)$P(7,3) lies outside $C$C. Find the equation of one of the tangents from $P$P to $C$C that is parallel to neither axis, giving your answer in the form $y = mx + c$y=mx+c. (5)

Solution with mark scheme:

(a) Step 1 — complete the square in $x$x and $y$y.

$x^2 - 6x = (x - 3)^2 - 9$x2−6x=(x−3)2−9 $y^2 + 4y = (y + 2)^2 - 4$y2+4y=(y+2)2−4

M1 — completing the square in both variables. The most common error here is a sign slip: writing $x^2 - 6x = (x - 3)^2 + 9$x2−6x=(x−3)2+9 (forgetting to subtract the squared half-coefficient) or $y^2 + 4y = (y - 2)^2 - 4$y2+4y=(y−2)2−4 (sign of the half-coefficient inside the bracket).

Step 2 — assemble.

$(x - 3)^2 - 9 + (y + 2)^2 - 4 - 12 = 0$(x−3)2−9+(y+2)2−4−12=0 $(x - 3)^2 + (y + 2)^2 = 25$(x−3)2+(y+2)2=25

A1 — correct form with $r^2 = 25$r2=25, not $r = 25$r=25.

A1 — centre $(3, -2)$(3,−2) and radius $r = 5$r=5, both stated explicitly.

(b) Step 1 — set up the tangent line through $P(7, 3)$P(7,3).

Let the tangent have gradient $m$m. Its equation is $y - 3 = m(x - 7)$y−3=m(x−7), i.e. $y = mx - 7m + 3$y=mx−7m+3.

M1 — writing the tangent as a one-parameter family through $P$P.

Step 2 — distance from centre equals radius.

The perpendicular distance from $(3, -2)$(3,−2) to the line $mx - y - 7m + 3 = 0$mx−y−7m+3=0 must equal $5$5:

$\dfrac{|3m - (-2) - 7m + 3|}{\sqrt{m^2 + 1}} = 5 \implies \dfrac{|{-4m + 5}|}{\sqrt{m^2 + 1}} = 5$m2+1​∣3m−(−2)−7m+3∣​=5⟹m2+1​∣−4m+5∣​=5

M1 — applying the distance formula with the tangency condition (distance from centre = radius). An equivalent route is to substitute into the circle equation and impose discriminant $= 0$=0; both earn the M1.

Step 3 — square both sides and solve.

$(-4m + 5)^2 = 25(m^2 + 1)$(−4m+5)2=25(m2+1) $16m^2 - 40m + 25 = 25m^2 + 25$16m2−40m+25=25m2+25 $-9m^2 - 40m = 0 \implies m(9m + 40) = 0$−9m2−40m=0⟹m(9m+40)=0

M1 — correct expansion and simplification to a quadratic in $m$m.

So $m = 0$m=0 (horizontal tangent — rejected by the question's "parallel to neither axis" instruction) or $m = -\dfrac{40}{9}$m=−940​.

A1 — correct gradient $m = -\tfrac{40}{9}$m=−940​ identified as the non-axis-parallel tangent.

Step 4 — write the tangent.

$y = -\dfrac{40}{9}x + \dfrac{280}{9} + 3 = -\dfrac{40}{9}x + \dfrac{307}{9}$y=−940​x+9280​+3=−940​x+9307​

A1 — final equation in the requested form $y = mx + c$y=mx+c.

Total: 8 marks (M3 A4 plus M1 setup, split as shown).

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Specimen question modelled on the AQA 7357 Paper 1 format

Question (6 marks): The circle $C$C has centre $(2, -1)$(2,−1) and passes through the point $A(5, 3)$A(5,3).

(a) Show that the equation of $C$C is $x^2 + y^2 - 4x + 2y - 20 = 0$x2+y2−4x+2y−20=0. (3)

(b) The line $\ell$ℓ has equation $y = 2x + k$y=2x+k for a constant $k$k. Find the values of $k$k for which $\ell$ℓ is a tangent to $C$C. (3)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — applying $r^2 = (5 - 2)^2 + (3 - (-1))^2 = 9 + 16 = 25$r2=(5−2)2+(3−(−1))2=9+16=25.
• M1 (AO1.1b) — writing $(x - 2)^2 + (y + 1)^2 = 25$(x−2)2+(y+1)2=25 and expanding.
• A1 (AO2.1) — printed form $x^2 + y^2 - 4x + 2y - 20 = 0$x2+y2−4x+2y−20=0 obtained, with $-20$−20 from $4 + 1 - 25$4+1−25.

(b)

• M1 (AO3.1a) — substituting $y = 2x + k$y=2x+k into the circle equation and reducing to a quadratic in $x$x (or equivalent — students may use the perpendicular-distance route instead).
• M1 (AO1.1b) — applying the discriminant-zero tangency condition or the distance-equals-radius condition.
• A1 (AO1.1b) — correct values of $k$k.

Total: 6 marks split AO1 = 4, AO2 = 1, AO3 = 1. AQA circle questions cluster heavily in AO1 with a single AO3 mark for choosing between substitution-discriminant and distance-from-centre approaches — both are accepted.

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Synoptic links

Connects to:

1. Section B — Algebra (completing the square): every "find centre and radius from general form" problem is a completing-the-square exercise dressed in coordinate-geometry clothing. The same algebra that finds the vertex of $y = x^2 - 6x + 5$y=x2−6x+5 finds the centre of the corresponding circle.

2. Section E — Trigonometry (parametric form): any point on the circle $(x - a)^2 + (y - b)^2 = r^2$(x−a)2+(y−b)2=r2 can be written $x = a + r\cos\theta$x=a+rcosθ, $y = b + r\sin\theta$y=b+rsinθ. This parametrisation is the bridge to parametric equations and to mechanics problems involving circular motion.

3. Section J — Differentiation (implicit): differentiating $(x - a)^2 + (y - b)^2 = r^2$(x−a)2+(y−b)2=r2 implicitly gives $2(x - a) + 2(y - b)\dfrac{dy}{dx} = 0$2(x−a)+2(y−b)dxdy​=0, hence $\dfrac{dy}{dx} = -\dfrac{x - a}{y - b}$dxdy​=−y−bx−a​. This recovers the tangent-perpendicular-to-radius result without geometric reasoning — a key synoptic check.

4. Section L — Vectors: the locus $|\mathbf{r} - \mathbf{c}| = r$∣r−c∣=r, where $\mathbf{r} = (x, y)$r=(x,y) and $\mathbf{c} = (a, b)$c=(a,b), is precisely the circle. This generalises to the sphere $|\mathbf{r} - \mathbf{c}| = r$∣r−c∣=r in 3D, and to the ball in higher dimensions.

5. GCSE circle theorems: the tangent-perpendicular-to-radius theorem from KS4 becomes the operational tool of the A-Level: tangent gradient $\times$× radius gradient $= -1$=−1. The "angle in a semicircle" theorem reappears when finding chords through opposite ends of a diameter.

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Mark-scheme literacy

AQA circle questions on Paper 1 split AO marks as follows:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	65–75%	Completing the square correctly, identifying centre and radius, applying the tangency condition
AO2 (reasoning / interpretation)	15–25%	Recognising when distance-from-centre is faster than substitution, justifying choice of method, presenting the equation in the requested form
AO3 (problem-solving)	5–15%	Choosing between Pythagoras and discriminant for a tangent-from-external-point problem; selecting the geometric route for an Apollonius-style locus