Circle Properties

Builds on: aqa-alevel-maths-pure-1 / coordinate-geometry — this lesson is the A2-level deep dive.

This lesson covers the key geometric properties of circles that are tested at A-Level. These properties connect algebra with geometry and are essential for solving problems involving tangents, chords, and intersections. You will use the equation of a circle together with these geometric facts to solve examination-style problems.

Spec Mapping — AQA 7357 Section C (Coordinate geometry). This lesson covers the circle-property content of Section C — tangents, chords, and angle-in-a-semicircle — at the depth required for A2-level synoptic questions. Refer to the official AQA specification document for exact wording.

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Property 1: The Tangent is Perpendicular to the Radius

At any point P on a circle, the tangent to the circle at P is perpendicular to the radius drawn from the centre to P.

This is the most frequently used circle property at A-Level.

Application: To find the equation of the tangent at a point P on a circle:

1. Find the gradient of the radius from the centre to P.
2. The tangent gradient is the negative reciprocal.
3. Use y − y₁ = m(x − x₁) with point P.

Example 1: The circle C has equation (x − 2)² + (y − 3)² = 20. Find the equation of the tangent at the point P(6, 5).

$$\begin{aligned} \text{Centre} &= (2,\ 3) \\[4pt] \text{Gradient of radius } CP &= \frac{5 - 3}{6 - 2} = \frac{2}{4} = \frac{1}{2} \\[4pt] \text{Gradient of tangent} &= -2 \quad \text{(negative reciprocal)} \\[6pt] \text{Tangent: } y - 5 &= -2(x - 6) \\ y - 5 &= -2x + 12 \\ y &= -2x + 17 \end{aligned}$$CentreGradient of radius CPGradient of tangentTangent: y−5y−5y​=(2, 3)=6−25−3​=42​=21​=−2(negative reciprocal)=−2(x−6)=−2x+12=−2x+17​

or equivalently, 2x + y − 17 = 0.

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Property 2: The Perpendicular from the Centre to a Chord Bisects the Chord

If a line is drawn from the centre of a circle perpendicular to a chord, then it bisects the chord (cuts it into two equal parts).

Conversely, the line from the centre to the midpoint of a chord is perpendicular to the chord.

Example 2: The circle C has centre (5, 4) and the chord AB has endpoints A(1, 1) and B(9, 7). Show that the perpendicular from the centre to AB bisects it.

$$\begin{aligned} &\text{Midpoint of } AB = \left( \frac{1 + 9}{2},\ \frac{1 + 7}{2} \right) = (5,\ 4) \\[4pt] &\text{The midpoint of } AB \text{ is the centre itself! Here the chord is a diameter.} \\[8pt] &\textbf{For a non-trivial example:} \\ &\text{Let } A(2, 6) \text{ and } B(8, 6) \text{ be a chord of a circle with centre } (5, 2). \\[4pt] &\text{Midpoint of } AB = (5,\ 6) \\[4pt] &\text{Gradient of } AB = \frac{6 - 6}{8 - 2} = 0 \quad \text{(horizontal chord)} \\[4pt] &\text{Line from centre } (5, 2) \text{ to midpoint } (5, 6) \text{ is vertical: } x = 5. \\[4pt] &\text{A vertical line is perpendicular to a horizontal line.} \;\checkmark \\ &\text{The perpendicular from the centre meets the chord at its midpoint.} \;\checkmark \end{aligned}$$​Midpoint of AB=(21+9​, 21+7​)=(5, 4)The midpoint of AB is the centre itself! Here the chord is a diameter.For a non-trivial example:Let A(2,6) and B(8,6) be a chord of a circle with centre (5,2).Midpoint of AB=(5, 6)Gradient of AB=8−26−6​=0(horizontal chord)Line from centre (5,2) to midpoint (5,6) is vertical: x=5.A vertical line is perpendicular to a horizontal line.✓The perpendicular from the centre meets the chord at its midpoint.✓​

Example 3: A chord PQ of the circle (x − 4)² + (y + 1)² = 50 has midpoint M(1, −4). Find the equation of PQ.

$$\begin{aligned} \text{Centre} &= (4,\ -1) \\[4pt] \text{Gradient of } CM &= \frac{-4 - (-1)}{1 - 4} = \frac{-3}{-3} = 1 \\[6pt] &\text{Since } CM \perp PQ: \\ \text{Gradient of } PQ &= -1 \\[6pt] PQ: y - (-4) &= -1(x - 1) \\ y + 4 &= -x + 1 \\ y &= -x - 3 \end{aligned}$$CentreGradient of CMGradient of PQPQ:y−(−4)y+4y​=(4, −1)=1−4−4−(−1)​=−3−3​=1Since CM⊥PQ:=−1=−1(x−1)=−x+1=−x−3​

or equivalently, x + y + 3 = 0.

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Property 3: Angle in a Semicircle

The angle in a semicircle is a right angle. That is, if AB is a diameter of a circle and P is any other point on the circle, then the angle APB = 90°.

In coordinate geometry, this means:

$$(\text{Gradient of } PA) \times (\text{Gradient of } PB) = -1$$(Gradient of PA)×(Gradient of PB)=−1

Example 4: A circle has diameter from A(−2, 0) to B(6, 4). The point P(2, 6) lies on the circle. Verify that angle APB = 90°.

$$\begin{aligned} &\text{Gradient of } PA = \frac{6 - 0}{2 - (-2)} = \frac{6}{4} = \frac{3}{2} \\[4pt] &\text{Gradient of } PB = \frac{6 - 4}{2 - 6} = \frac{2}{-4} = -\frac{1}{2} \\[4pt] &\text{Product} = \frac{3}{2} \times \left(-\frac{1}{2}\right) = -\frac{3}{4} \\[6pt] &\text{This is not } -1, \text{ so we must check whether } P \text{ lies on the circle.} \\[4pt] &\text{Centre} = \left( \frac{-2 + 6}{2},\ \frac{0 + 4}{2} \right) = (2,\ 2) \\[4pt] &\text{Radius}^{2} = (6 - 2)^{2} + (4 - 2)^{2} = 16 + 4 = 20 \\[4pt] &\text{Check } P(2, 6): (2 - 2)^{2} + (6 - 2)^{2} = 0 + 16 = 16 \neq 20 \\[4pt] &\text{So } P(2, 6) \text{ does not lie on this circle. Find a point that does.} \\[4pt] &\text{Let } P(6, 0): \quad (6 - 2)^{2} + (0 - 2)^{2} = 16 + 4 = 20 \;\checkmark \\[4pt] &\text{Gradient of } PA = \frac{0 - 0}{6 - (-2)} = \frac{0}{8} = 0 \\[4pt] &\text{Gradient of } PB = \frac{0 - 4}{6 - 6} \to \text{undefined (vertical)} \\[4pt] &\text{A horizontal line is perpendicular to a vertical line, so } \angle APB = 90^{\circ}. \;\checkmark \end{aligned}$$​Gradient of PA=2−(−2)6−0​=46​=23​Gradient of PB=2−66−4​=−42​=−21​Product=23​×(−21​)=−43​This is not −1, so we must check whether P lies on the circle.Centre=(2−2+6​, 20+4​)=(2, 2)Radius2=(6−2)2+(4−2)2=16+4=20Check P(2,6):(2−2)2+(6−2)2=0+16=16=20So P(2,6) does not lie on this circle. Find a point that does.Let P(6,0):(6−2)2+(0−2)2=16+4=20✓Gradient of PA=6−(−2)0−0​=80​=0Gradient of PB=6−60−4​→undefined (vertical)A horizontal line is perpendicular to a vertical line, so ∠APB=90∘.✓​

More general example: Let A(0, 0) and B(8, 0) be a diameter. Centre = (4, 0), radius = 4. Point P(4, 4) lies on the circle since (4−4)² + (4−0)² = 16 = r².

$$\begin{aligned} \text{Gradient of } PA &= 4/4 = 1 \\ \text{Gradient of } PB &= (4 - 0)/(4 - 8) = 4/(-4) = -1 \\ \text{Product} &= 1 \times (-1) = -1 \;\checkmark \end{aligned}$$Gradient of PAGradient of PBProduct​=4/4=1=(4−0)/(4−8)=4/(−4)=−1=1×(−1)=−1✓​

Angle APB = 90°, confirming the angle-in-a-semicircle theorem.

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Finding Where a Line Intersects a Circle

To find the intersection points of a straight line and a circle, substitute the equation of the line into the equation of the circle and solve the resulting quadratic.

The discriminant determines the number of intersection points:

• $\Delta > 0$Δ>0: two intersection points (the line is a secant).
• $\Delta = 0$Δ=0: one intersection point (the line is a tangent).
• $\Delta < 0$Δ<0: no intersection points (the line misses the circle).

Example 5: Find where the line y = x + 1 meets the circle x² + y² = 25.

$$\begin{aligned} &\text{Substitute } y = x + 1: \\ x^{2} + (x + 1)^{2} &= 25 \\ x^{2} + x^{2} + 2x + 1 &= 25 \\ 2x^{2} + 2x - 24 &= 0 \\ x^{2} + x - 12 &= 0 \\ (x + 4)(x - 3) &= 0 \\ x = -4 \ &\text{or}\ x = 3 \\[6pt] \text{When } x = -4: y = -3 \quad &\text{When } x = 3: y = 4 \end{aligned}$$x2+(x+1)2x2+x2+2x+12x2+2x−24x2+x−12(x+4)(x−3)x=−4 When x=−4:y=−3​Substitute y=x+1:=25=25=0=0=0or x=3When x=3:y=4​

Intersection points: (−4, −3) and (3, 4).

Example 6: Show that the line y = 3x + 10 is a tangent to the circle x² + y² = 10.

$$\begin{aligned} &\text{Substitute:} \\ x^{2} + (3x + 10)^{2} &= 10 \\ x^{2} + 9x^{2} + 60x + 100 &= 10 \\ 10x^{2} + 60x + 90 &= 0 \\ x^{2} + 6x + 9 &= 0 \\ (x + 3)^{2} &= 0 \\ x &= -3 \quad \text{(repeated root)} \end{aligned}$$x2+(3x+10)2x2+9x2+60x+10010x2+60x+90x2+6x+9(x+3)2x​Substitute:=10=10=0=0=0=−3(repeated root)​

Since the discriminant is zero (repeated root), the line is a tangent. The point of tangency is (−3, 1).

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Finding the Length of a Tangent from an External Point

The length of a tangent from an external point $(x_1, y_1)$(x1​,y1​) to a circle with centre $(a, b)$(a,b) and radius $r$r can be found using Pythagoras' theorem:

$$\text{tangent length} = \sqrt{d^{2} - r^{2}}$$tangent length=d2−r2​

where d is the distance from the external point to the centre.

Example 7: Find the length of the tangent from the point (7, 4) to the circle (x − 1)² + (y − 2)² = 9.

$$\begin{aligned} \text{Centre} &= (1,\ 2), \quad \text{radius} = 3 \\[4pt] d^{2} &= (7 - 1)^{2} + (4 - 2)^{2} = 36 + 4 = 40 \\[4pt] \text{Tangent length} &= \sqrt{40 - 9} = \sqrt{31} \end{aligned}$$Centred2Tangent length​=(1, 2),radius=3=(7−1)2+(4−2)2=36+4=40=40−9​=31​​

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Specimen Question (Modelled on the AQA Paper Format)

Note: this question is constructed to model AQA Paper 1/2/3 style; it is not a reproduction of any published past paper.

Worked Examination-Style Problem

Problem: The circle C has equation x² + y² − 10x − 4y + 20 = 0. The line l has equation y = kx, where k is a constant.

(a) Find the centre and radius of C.

(b) Find the values of k for which l is a tangent to C.

Solution:

(a)

$$\begin{aligned} (x^{2} - 10x) + (y^{2} - 4y) &= -20 \\ (x - 5)^{2} - 25 + (y - 2)^{2} - 4 &= -20 \\ (x - 5)^{2} + (y - 2)^{2} &= 9 \\[4pt] \text{Centre} = (5,\ 2), \quad &\text{radius} = 3 \end{aligned}$$(x2−10x)+(y2−4y)(x−5)2−25+(y−2)2−4(x−5)2+(y−2)2Centre=(5, 2),​=−20=−20=9radius=3​

(b) Substitute y = kx:

$$\begin{aligned} x^{2} + k^{2}x^{2} - 10x - 4kx + 20 &= 0 \\ (1 + k^{2})x^{2} - (10 + 4k)x + 20 &= 0 \\[6pt] &\text{For tangency, discriminant} = 0: \\ (10 + 4k)^{2} - 4(1 + k^{2})(20) &= 0 \\ 100 + 80k + 16k^{2} - 80 - 80k^{2} &= 0 \\ -64k^{2} + 80k + 20 &= 0 \\ 64k^{2} - 80k - 20 &= 0 \\ 16k^{2} - 20k - 5 &= 0 \\[4pt] k &= \frac{20 \pm \sqrt{400 + 320}}{32} \\ k &= \frac{20 \pm \sqrt{720}}{32} \\ k &= \frac{20 \pm 12\sqrt{5}}{32} \\ k &= \frac{5 \pm 3\sqrt{5}}{8} \end{aligned}$$x2+k2x2−10x−4kx+20(1+k2)x2−(10+4k)x+20(10+4k)2−4(1+k2)(20)100+80k+16k2−80−80k2−64k2+80k+2064k2−80k−2016k2−20k−5kkkk​=0=0For tangency, discriminant=0:=0=0=0=0=0=3220±400+320​​=3220±720​​=3220±125​​=85±35​​​

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Synoptic Links

This topic connects to:

• Equation of a circle (aqa-alevel-maths-coordinate-geometry-depth / equation-of-a-circle) — the algebraic form on which all these geometric properties are exploited.
• Straight lines: review & extension (aqa-alevel-maths-coordinate-geometry-depth / straight-lines-review) — gradients and perpendicularity supply the tangent and chord-bisector machinery.
• Loci and coordinate geometry proofs (aqa-alevel-maths-coordinate-geometry-depth / loci-and-geometric-proof) — angle-in-a-semicircle and perpendicular-bisector arguments are core coordinate-proof techniques.

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Summary

• The tangent at any point on a circle is perpendicular to the radius at that point.
• The perpendicular from the centre to a chord bisects the chord.
• The angle in a semicircle is 90°.
• To find line-circle intersections, substitute and solve the resulting quadratic.
• The discriminant tells you: Δ > 0 (secant, two points), Δ = 0 (tangent), Δ < 0 (no intersection).
• Tangent length from an external point: $\sqrt{d^2 - r^2}$d2−r2​.

Exam Tip: When finding where a line meets a circle, always simplify the quadratic fully before attempting to solve it. If the question says "show that the line is a tangent", you must show that the discriminant equals zero — do not just state it. For tangent equations, always begin by finding the gradient of the radius to the point of tangency.

A-Level Deep Dive: Circle Properties

Spec mapping

AQA 7357 specification, Paper 1 — Pure Mathematics, Section C: Coordinate Geometry, sub-strand on circles. The specification requires understanding and use of: the equation of a circle in the form $(x - a)^2 + (y - b)^2 = r^2$(x−a)2+(y−b)2=r2; the geometric properties that the tangent at any point on a circle is perpendicular to the radius at that point; that the perpendicular from the centre of a circle to a chord bisects the chord; and the converse, that the angle in a semicircle is a right angle. Although introduced in Section C, these properties are examined synoptically across Paper 1 (Pure), Paper 2 (Pure and Mechanics) — where loci/path arguments use circle geometry — and Paper 3 (Pure and Statistics) through coordinate-based modelling. The AQA formula booklet provides the general circle equation but does not state the three geometric theorems — they must be known and quoted.

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Worked example with full mark scheme

Question (8 marks):

A circle $C$C has equation $x^2 + y^2 - 6x + 4y - 12 = 0$x2+y2−6x+4y−12=0. The point $P(9, 1)$P(9,1) lies outside $C$C.

(a) Find the centre and radius of $C$C. (2)

(b) The chord $AB$AB has midpoint $M(3, 2)$M(3,2). Find the perpendicular distance from the centre of $C$C to $AB$AB, and hence the length of $AB$AB. (3)

(c) Find the length of the tangent from $P$P to $C$C. (3)

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 and $y^2 + 4y = (y + 2)^2 - 4$y2+4y=(y+2)2−4, so the equation becomes $(x - 3)^2 - 9 + (y + 2)^2 - 4 - 12 = 0$(x−3)2−9+(y+2)2−4−12=0, i.e. $(x - 3)^2 + (y + 2)^2 = 25$(x−3)2+(y+2)2=25.

M1 — completing the square in both variables.

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

(b) Step 1 — apply the chord-bisection theorem.

The perpendicular from the centre $C(3, -2)$C(3,−2) to a chord meets the chord at its midpoint. So the perpendicular distance equals $|CM|$∣CM∣:

$|CM| = \sqrt{(3 - 3)^2 + (2 - (-2))^2} = \sqrt{0 + 16} = 4$∣CM∣=(3−3)2+(2−(−2))2​=0+16​=4

M1 — quoting and using the perpendicular-bisector theorem.

Step 2 — apply Pythagoras for chord length.

Triangle $CMA$CMA (where $A$A is an endpoint of the chord) is right-angled at $M$M, with hypotenuse $CA = r = 5$CA=r=5 and one leg $CM = 4$CM=4.

$MA^2 = r^2 - CM^2 = 25 - 16 = 9 \implies MA = 3$MA2=r2−CM2=25−16=9⟹MA=3

Hence $AB = 2 \cdot MA = 6$AB=2⋅MA=6.

A1 — perpendicular distance $4$4, chord length $6$6.

B1 — both values stated with the chord-bisection reason explicit.

(c) Step 1 — distance from $P$P to centre.

$CP = \sqrt{(9 - 3)^2 + (1 - (-2))^2} = \sqrt{36 + 9} = \sqrt{45}$CP=(9−3)2+(1−(−2))2​=36+9​=45​

M1 — applying the distance formula to find $CP$CP.

Step 2 — apply tangent-length formula.

Because the tangent at the point of contact $T$T is perpendicular to the radius $CT$CT, triangle $CTP$CTP is right-angled at $T$T. By Pythagoras:

$PT^2 = CP^2 - r^2 = 45 - 25 = 20$PT2=CP2−r2=45−25=20

M1 — using $PT = \sqrt{CP^2 - r^2}$PT=CP2−r2​, citing the tangent-perpendicular-to-radius property.

A1 — $PT = \sqrt{20} = 2\sqrt{5}$PT=20​=25​.

Total: 8 marks (M4 A3 B1).

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Specimen question modelled on AQA 7357 Paper 1 format

Question (6 marks): The points $A(2, 6)$A(2,6) and $B(8, -2)$B(8,−2) are the endpoints of a diameter of a circle $C$C.

(a) State, with reason, the size of the angle $\angle APB$∠APB for any point $P$P on the circle (other than $A$A or $B$B). (2)

(b) Find the equation of $C$C in the form $(x - a)^2 + (y - b)^2 = r^2$(x−a)2+(y−b)2=r2. (4)

Mark scheme decomposition by AO:

(a)

• B1 (AO1.2) — $\angle APB = 90°$∠APB=90°.
• B1 (AO2.4) — reason: angle in a semicircle is a right angle (since $AB$AB is a diameter).

(b)

• M1 (AO1.1a) — centre is the midpoint of the diameter: $\left(\dfrac{2 + 8}{2}, \dfrac{6 + (-2)}{2}\right) = (5, 2)$(22+8​,26+(−2)​)=(5,2).
• M1 (AO1.1b) — radius is half the diameter: $AB = \sqrt{(8 - 2)^2 + (-2 - 6)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$AB=(8−2)2+(−2−6)2​=36+64​=100​=10, so $r = 5$r=5.
• A1 (AO1.1b) — $r^2 = 25$r2=25.
• A1 (AO2.5) — equation $(x - 5)^2 + (y - 2)^2 = 25$(x−5)2+(y−2)2=25.

Total: 6 marks split AO1 = 4, AO2 = 2. This problem rewards recognition that "endpoints of a diameter" is the key phrase: the midpoint gives the centre and the half-distance gives the radius without needing the general implicit form.

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Synoptic links

Connects to:

1. Section C — Equation of a circle (general implicit form): completing the square converts $x^2 + y^2 + Dx + Ey + F = 0$x2+y2+Dx+Ey+F=0 to centre-radius form $(x - a)^2 + (y - b)^2 = r^2$(x−a)2+(y−b)2=r2. Every circle-properties problem requires this conversion as a first step when the implicit form is given.

2. Section B — Algebra and functions: finding intersections of lines and circles produces a quadratic; the discriminant determines whether the line is a tangent ($\Delta = 0$Δ=0), secant ($\Delta > 0$Δ>0) or non-intersecting ($\Delta < 0$Δ<0). The tangent-perpendicular-to-radius property gives an alternative, often quicker, route to tangent equations.

3. Section J — Vectors: the tangent-radius perpendicularity is exactly the vector condition $\vec{CT} \cdot \vec{TP} = 0$CT⋅TP=0. Many circle problems accept either a coordinate-geometry or a vector approach, and vector dot-product cleanly handles 3D extensions.

4. Section E — Trigonometry (sector and segment areas): the chord-bisection theorem produces the half-angle subtended at the centre, allowing computation of segment areas via $\tfrac{1}{2}r^2(\theta - \sin\theta)$21​r2(θ−sinθ) once the chord length and radius are known.