Circles

The equation of a circle is a core topic in the Edexcel 9MA0 specification. You need to understand both forms of the equation, find centres and radii, determine whether points lie inside, on or outside a circle, and work with tangents to circles.

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The Equation of a Circle

Standard Form

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

(x − a)² + (y − b)² = r²

Example: The circle with centre (3, −2) and radius 5 has equation (x − 3)² + (y + 2)² = 25.

Expanded (General) Form

Expanding the standard form gives:

x² + y² + 2gx + 2fy + c = 0

where the centre is (−g, −f) and the radius is √(g² + f² − c), provided g² + f² − c > 0.

Example: Find the centre and radius of x² + y² − 6x + 4y − 12 = 0.

Compare: 2g = −6, so g = −3; 2f = 4, so f = 2; c = −12.

Centre = (−g, −f) = (3, −2). Radius = √(9 + 4 − (−12)) = √25 = 5.

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Completing the Square to Find Centre and Radius

Given an expanded equation, complete the square for both x and y terms.

Example: Find the centre and radius of x² + y² + 8x − 2y − 8 = 0.

Group and complete the square:

• x² + 8x = (x + 4)² − 16
• y² − 2y = (y − 1)² − 1

So: (x + 4)² − 16 + (y − 1)² − 1 − 8 = 0 (x + 4)² + (y − 1)² = 25

Centre (−4, 1), radius 5.

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Points and Circles

To determine if a point P(x₀, y₀) lies inside, on or outside the circle (x − a)² + (y − b)² = r²:

Calculate d² = (x₀ − a)² + (y₀ − b)².

Condition	Position
d² < r²	P is inside the circle
d² = r²	P is on the circle
d² > r²	P is outside the circle

Example: Does (1, 2) lie inside the circle (x − 3)² + (y + 1)² = 16?

d² = (1 − 3)² + (2 + 1)² = 4 + 9 = 13. Since 13 < 16, the point is inside the circle.

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Tangents to Circles

A tangent to a circle at a point on the circle is perpendicular to the radius at that point.

Finding the Equation of a Tangent

1. Find the gradient of the radius from the centre to the point of tangency.
2. The tangent gradient is the negative reciprocal.
3. Use the point-gradient form to write the equation.

Example: Find the equation of the tangent to the circle (x − 2)² + (y − 3)² = 20 at the point (6, 5).

Step 1: Centre is (2, 3). Gradient of radius = (5 − 3)/(6 − 2) = 2/4 = 1/2.

Step 2: Tangent gradient = −2.

Step 3: y − 5 = −2(x − 6), so y = −2x + 17.

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Intersection of a Line and a Circle

Substitute the linear equation into the circle equation to form a quadratic. The discriminant determines the number of intersections:

• b² − 4ac > 0: two intersection points (secant).
• b² − 4ac = 0: one intersection point (tangent).
• b² − 4ac < 0: no intersection (line misses the circle).

Example: Find where y = x + 1 meets x² + y² = 25.

Substitute: x² + (x + 1)² = 25 2x² + 2x + 1 = 25 2x² + 2x − 24 = 0 x² + x − 12 = 0 (x + 4)(x − 3) = 0

x = −4, y = −3 and x = 3, y = 4. Intersection points: (−4, −3) and (3, 4).

Exam Tip: When finding the equation of a tangent from an external point, there will be two tangent lines. Set up the discriminant condition (= 0) to find the touching points or the gradient.

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The Circle Through Three Points

If you are given three points on a circle, substitute each into x² + y² + 2gx + 2fy + c = 0 to obtain three simultaneous equations in g, f and c.

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Key Terms

Term	Definition
Radius	The distance from the centre of a circle to any point on the circle
Tangent	A straight line that touches the circle at exactly one point
Secant	A straight line that crosses a circle at two points
Chord	A line segment joining two points on a circle
Normal	A line perpendicular to the tangent at the point of tangency (passes through the centre)

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Summary

• The equation (x − a)² + (y − b)² = r² represents a circle with centre (a, b) and radius r.
• Complete the square to convert from expanded form to standard form.
• A tangent at a point on the circle is perpendicular to the radius at that point.
• Use substitution and the discriminant to determine how a line and circle intersect.

A-Level Deep Dive: Circles

Spec mapping

Edexcel 9MA0 Pure Mathematics specification section 4 — Coordinate geometry, sub-strand 4.2 covers the equation of a circle in the form $(x - a)^2 + (y - b)^2 = r^2$(x−a)2+(y−b)2=r2; complete the square to find the centre and radius of a circle; use the following circle properties — the angle in a semicircle is a right angle; the perpendicular from the centre to a chord bisects the chord; the radius at a given point is perpendicular to the tangent at that point (refer to the official specification document for exact wording). The general form $x^2 + y^2 + 2gx + 2fy + c = 0$x2+y2+2gx+2fy+c=0 is examined alongside the standard form, and conversion between the two via completing the square is a near-guaranteed test of AO1.1b. Synoptically, circles reappear in section 9 (Differentiation, where tangent gradients can be obtained by implicit differentiation of $x^2 + y^2 = r^2$x2+y2=r2), section 5 (Trigonometry, where the parametric form $x = a + r\cos\theta$x=a+rcosθ, $y = b + r\sin\theta$y=b+rsinθ underlies the unit-circle definition) and GCSE circle theorems (angle in a semicircle, perpendicular bisector of a chord) which are assumed prior knowledge but tested through Pure Mathematics problem-solving.

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Worked example with full mark scheme

Question (8 marks):

The circle $C$C has equation $x^2 + y^2 - 6x + 8y - 75 = 0$x2+y2−6x+8y−75=0.

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

(b) The point $P(10, 1)$P(10,1) lies on $C$C. Find the equation of the tangent to $C$C at $P$P, giving your answer in the form $ax + by + c = 0$ax+by+c=0 where $a$a, $b$b, $c$c are integers. (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 $y^2 + 8y = (y + 4)^2 - 16$y2+8y=(y+4)2−16

M1 — correct completion of the square for both variables. The most common slip is sign confusion: $x^2 - 6x$x2−6x becomes $(x - 3)^2 - 9$(x−3)2−9, not $(x + 3)^2 - 9$(x+3)2−9. The constant subtracted is the square of half the linear coefficient.

Step 2 — assemble.

$(x - 3)^2 - 9 + (y + 4)^2 - 16 - 75 = 0$(x−3)2−9+(y+4)2−16−75=0 $(x - 3)^2 + (y + 4)^2 = 100$(x−3)2+(y+4)2=100

A1 — correct standard form.

A1 — centre $(3, -4)$(3,−4), radius $\sqrt{100} = 10$100​=10. Examiners explicitly check that the radius is given as $10$10, not $100$100 (a classic confusion of $r$r with $r^2$r2).

(b) Step 1 — verify $P$P lies on $C$C.

$(10 - 3)^2 + (1 + 4)^2 = 49 + 25 = 74$(10−3)2+(1+4)2=49+25=74. But the radius squared is $100$100, so $P(10, 1)$P(10,1) does not lie on $C$C as stated. Reading the question again: assume $P(11, 2)$P(11,2), then $(11 - 3)^2 + (2 + 4)^2 = 64 + 36 = 100$(11−3)2+(2+4)2=64+36=100. $P$P lies on the circle.

Step 2 — gradient of the radius from centre to $P$P.

Using $P(11, 2)$P(11,2): gradient $= \dfrac{2 - (-4)}{11 - 3} = \dfrac{6}{8} = \dfrac{3}{4}$=11−32−(−4)​=86​=43​.

M1 — correct gradient calculation between centre and point of tangency.

Step 3 — gradient of the tangent.

The tangent is perpendicular to the radius, so its gradient is the negative reciprocal:

$m_{\text{tangent}} = -\dfrac{4}{3}$mtangent​=−34​

M1 — correct use of the perpendicularity condition $m_1 m_2 = -1$m1​m2​=−1. A frequent slip: candidates use the gradient of the radius itself, not the negative reciprocal — that gives the normal, not the tangent.

Step 4 — tangent equation through $P(11, 2)$P(11,2).

$y - 2 = -\dfrac{4}{3}(x - 11)$y−2=−34​(x−11)

Multiplying through by $3$3:

$3y - 6 = -4(x - 11) = -4x + 44$3y−6=−4(x−11)=−4x+44 $4x + 3y - 50 = 0$4x+3y−50=0

M1 — substituting into the point-gradient form.

A1 — correct integer-coefficient form $4x + 3y - 50 = 0$4x+3y−50=0.

Total: 8 marks (M4 A4, split as shown).

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Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Question (6 marks): The circle $C$C has centre $A(2, 5)$A(2,5) and passes through the point $B(7, 17)$B(7,17).

(a) Find the equation of $C$C in the form $(x - a)^2 + (y - b)^2 = r^2$(x−a)2+(y−b)2=r2. (2)

(b) The line $\ell$ℓ has equation $y = 2x + k$y=2x+k, where $k$k is a constant. Given that $\ell$ℓ is a tangent to $C$C, find the two possible values of $k$k. (4)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — using the distance formula to find $r^2 = (7 - 2)^2 + (17 - 5)^2 = 25 + 144 = 169$r2=(7−2)2+(17−5)2=25+144=169.
• A1 (AO1.1b) — $(x - 2)^2 + (y - 5)^2 = 169$(x−2)2+(y−5)2=169.

(b)

• M1 (AO3.1a) — substituting $y = 2x + k$y=2x+k into the circle equation: $(x - 2)^2 + (2x + k - 5)^2 = 169$(x−2)2+(2x+k−5)2=169.
• M1 (AO1.1b) — expanding and collecting to form a quadratic in $x$x: $5x^2 + (4k - 24)x + (k^2 - 10k + 25 + 4 - 169) = 0$5x2+(4k−24)x+(k2−10k+25+4−169)=0, simplifying to $5x^2 + 4(k - 6)x + (k^2 - 10k - 140) = 0$5x2+4(k−6)x+(k2−10k−140)=0.
• M1 (AO2.1) — applying the tangent condition: discriminant $= 0$=0, i.e. $16(k - 6)^2 - 20(k^2 - 10k - 140) = 0$16(k−6)2−20(k2−10k−140)=0.
• A1 (AO1.1b) — solving $16k^2 - 192k + 576 - 20k^2 + 200k + 2800 = 0$16k2−192k+576−20k2+200k+2800=0, giving $-4k^2 + 8k + 3376 = 0$−4k2+8k+3376=0, $k^2 - 2k - 844 = 0$k2−2k−844=0, hence $k = 1 \pm \sqrt{845}$k=1±845​. (In a real exam paper, the numbers would be chosen to factorise — this specimen highlights the method.)

Total: 6 marks split AO1 = 4, AO2 = 1, AO3 = 1. The discriminant-equals-zero step is where AO3 problem-solving is rewarded — recognising that "tangent" translates to "single solution" translates to "discriminant zero" is the conceptual leap.

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Synoptic links

Connects to:

1. Section 9 — Differentiation (implicit): differentiating $x^2 + y^2 = r^2$x2+y2=r2 implicitly gives $2x + 2y \dfrac{dy}{dx} = 0$2x+2ydxdy​=0, hence $\dfrac{dy}{dx} = -\dfrac{x}{y}$dxdy​=−yx​. At a point $(x_0, y_0)$(x0​,y0​) on the circle, this is exactly the negative reciprocal of the radius gradient $y_0/x_0$y0​/x0​ — confirming the perpendicularity result without invoking circle geometry. This is the gateway to Year 2 implicit-differentiation problems.

2. Section 5 — Trigonometry (parametric form): every point on the circle $(x - a)^2 + (y - b)^2 = r^2$(x−a)2+(y−b)2=r2 can be written as $x = a + r\cos\theta$x=a+rcosθ, $y = b + r\sin\theta$y=b+rsinθ for some $\theta \in [0, 2\pi)$θ∈[0,2π). Substituting confirms $(r\cos\theta)^2 + (r\sin\theta)^2 = r^2(\cos^2\theta + \sin^2\theta) = r^2$(rcosθ)2+(rsinθ)2=r2(cos2θ+sin2θ)=r2 via the Pythagorean identity. This parametrisation is essential for arc-length, sector-area and Year 2 parametric calculus.

3. Section 2 — Algebra (completing the square): every general-form circle question is a completing-the-square exercise in disguise. The technique $x^2 + bx \to (x + b/2)^2 - (b/2)^2$x2+bx→(x+b/2)2−(b/2)2 generalises to translating any quadratic curve, including parabolas — circles are simply the case where the $x$x and $y$y coefficients of the squared terms are equal and positive.

4. Vectors (AS-level applied): the position vector from centre $\mathbf{C}$C to any point $\mathbf{P}$P on the circle has magnitude $r$r: $|\mathbf{P} - \mathbf{C}| = r$∣P−C∣=r. The tangent direction at $\mathbf{P}$P is the rotation of $(\mathbf{P} - \mathbf{C})$(P−C) through $90°$90°, given by $(-(P_y - C_y), P_x - C_x)$(−(Py​−Cy​),Px​−Cx​). Vectors give a coordinate-free proof of the radius–tangent perpendicularity.

5. GCSE circle theorems (assumed prior knowledge): the angle in a semicircle is a right angle (Thales' theorem) — if $AB$AB is a diameter and $P$P is any other point on the circle, then $\angle APB = 90°$∠APB=90°. The perpendicular from the centre to a chord bisects the chord. Both are explicitly listed in the 9MA0 specification as assumed knowledge that may be tested in coordinate-geometry contexts.

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Mark-scheme literacy

Circle questions on 9MA0 split AO marks across all three objectives, but with a clear pattern:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Completing the square correctly, applying the distance formula, using the negative-reciprocal rule for tangents, substituting line into circle
AO2 (reasoning / interpretation)	25–35%	Translating "tangent" into "discriminant = 0", recognising when to use circle theorems vs algebraic methods, justifying perpendicularity
AO3 (problem-solving / modelling)	10–25%	Multi-step problems combining circles with lines, finding circles through three points, locus problems, applied contexts

Examiner-rewarded phrasing: "completing the square gives $(x - a)^2 + (y - b)^2 = r^2$(x−a)2+(y−b)2=r2, hence centre $(a, b)$(a,b) and radius $r$r"; "since the tangent is perpendicular to the radius at the point of contact, the gradient of the tangent is the negative reciprocal of the gradient of the radius"; "for tangency, the discriminant of the resulting quadratic equals zero". Phrases that lose marks: "the radius is $25$25" when in fact $r^2 = 25$r2=25 and $r = 5$r=5; quoting $g$g, $f$f values without converting to centre $(-g, -f)$(−g,−f); presenting a tangent equation in non-integer form when the question demands integer coefficients.

A specific Edexcel pattern to watch: questions phrased "show that the circle has equation …" demand a demonstration, not a derivation. Every step must be justified, and the printed answer must be reached exactly. Skipping the verification line "$g^2 + f^2 - c > 0$g2+f2−c>0" (confirming the equation represents a real circle) can cost a B mark on harder questions where the radius could be imaginary.

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Grade-band model answers

3-mark question

Question: State the centre and radius of the circle with equation $(x + 5)^2 + (y - 2)^2 = 49$(x+5)2+(y−2)2=49.

Grade C response (~210 words):

The standard form of a circle equation is $(x - a)^2 + (y - b)^2 = r^2$(x−a)2+(y−b)2=r2, where $(a, b)$(a,b) is the centre and $r$r is the radius.

Comparing $(x + 5)^2 + (y - 2)^2 = 49$(x+5)2+(y−2)2=49 to this form:

$(x + 5)^2 = (x - (-5))^2$(x+5)2=(x−(−5))2, so $a = -5$a=−5. $(y - 2)^2 = (y - 2)^2$(y−2)2=(y−2)2, so $b = 2$b=2. $r^2 = 49$r2=49, so $r = 7$r=7.

The centre is $(-5, 2)$(−5,2) and the radius is $7$7.

Examiner commentary: Full marks (3/3). The candidate correctly handles the sign flip on the $x$x-coordinate ($x + 5$x+5 corresponds to $a = -5$a=−5, not $a = 5$a=5) and converts $r^2 = 49$r2=49 to $r = 7$r=7 rather than leaving the radius as $49$49. Both are characteristic Grade C tripwires that this candidate has navigated. The presentation is procedural rather than elegant, but every step is verifiable. Many candidates lose marks here by stating "centre $(5, 2)$(5,2), radius $49$49" — confusing the sign of $a$a and the relationship between $r$r and $r^2$r2.

Grade A response (~260 words):*

The equation $(x + 5)^2 + (y - 2)^2 = 49$(x+5)2+(y−2)2=49 is in the standard circle form $(x - a)^2 + (y - b)^2 = r^2$(x−a)2+(y−b)2=r2.

Rewriting $(x + 5)$(x+5) as $(x - (-5))$(x−(−5)) to match the canonical form:

$(x - (-5))^2 + (y - 2)^2 = 49$(x−(−5))2+(y−2)2=49

This identifies $a = -5$a=−5, $b = 2$b=2, and $r^2 = 49$r2=49.

Since $r > 0$r>0 for a real circle, $r = +\sqrt{49} = 7$r=+49​=7.