Circle Properties

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.

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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).

Centre = (2, 3)

Gradient of radius CP = (5 − 3)/(6 − 2) = 2/4 = 1/2

Gradient of tangent = −2  (negative reciprocal)

Tangent: y − 5 = −2(x − 6)
         y − 5 = −2x + 12
         y = −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.

Midpoint of AB = ((1 + 9)/2, (1 + 7)/2) = (5, 4)

The midpoint of AB is the centre itself! In this case 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 = (6 − 6)/(8 − 2) = 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.

Centre = (4, −1)

Gradient of CM = (−4 − (−1))/(1 − 4) = (−3)/(−3) = 1

Since CM is perpendicular to PQ:
Gradient of PQ = −1

PQ: y − (−4) = −1(x − 1)
    y + 4 = −x + 1
    y = −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:

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°.

Gradient of PA = (6 − 0)/(2 − (−2)) = 6/4 = 3/2
Gradient of PB = (6 − 4)/(2 − 6) = 2/(−4) = −1/2

Product = (3/2) × (−1/2) = −3/4

Wait — this is not −1, so either P does not lie on the circle or we must check.

Centre = ((−2+6)/2, (0+4)/2) = (2, 2)
Radius² = (6 − 2)² + (4 − 2)² = 16 + 4 = 20

Check P(2, 6): (2−2)² + (6−2)² = 0 + 16 = 16 ≠ 20

So P(2, 6) does not lie on this circle. Let us find a point that does.

Let P(6, 0):
Check: (6−2)² + (0−2)² = 16 + 4 = 20 ✓

Gradient of PA = (0 − 0)/(6 − (−2)) = 0/8 = 0
Gradient of PB = (0 − 4)/(6 − 6) → undefined (vertical)

A horizontal line is perpendicular to a vertical line, so angle 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.

Gradient of PA = 4/4 = 1
Gradient of PB = (4 − 0)/(4 − 8) = 4/(−4) = −1
Product = 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.

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

When x = −4: y = −3    When 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.

Substitute:
x² + (3x + 10)² = 10
x² + 9x² + 60x + 100 = 10
10x² + 60x + 90 = 0
x² + 6x + 9 = 0
(x + 3)² = 0
x = −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:

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.

Centre = (1, 2), radius = 3

d² = (7 − 1)² + (4 − 2)² = 36 + 4 = 40

Tangent length = √(40 − 9) = √31

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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)

(x² − 10x) + (y² − 4y) = −20
(x − 5)² − 25 + (y − 2)² − 4 = −20
(x − 5)² + (y − 2)² = 9

Centre = (5, 2), radius = 3

(b) Substitute y = kx:

x² + k²x² − 10x − 4kx + 20 = 0
(1 + k²)x² − (10 + 4k)x + 20 = 0

For tangency, discriminant = 0:
(10 + 4k)² − 4(1 + k²)(20) = 0
100 + 80k + 16k² − 80 − 80k² = 0
−64k² + 80k + 20 = 0
64k² − 80k − 20 = 0
16k² − 20k − 5 = 0

k = (20 ± √(400 + 320)) / 32
k = (20 ± √720) / 32
k = (20 ± 12√5) / 32
k = (5 ± 3√5) / 8

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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: √(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.

5. Section A — Proof: the three named theorems (tangent-radius perpendicularity, perpendicular-from-centre bisects chord, angle in semicircle) are all proved by coordinate or vector argument. AQA increasingly examines geometric proof at A-Level, and these are canonical examples.

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Mark-scheme literacy

Circle-properties questions on AQA 7357 split AO marks as follows:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	55–65%	Completing the square; applying distance formula; using Pythagoras with radius / chord half-length / perpendicular distance
AO2 (reasoning / interpretation)	25–35%	Stating the named theorem being used; justifying tangent length via right-angled triangle; recognising "diameter" implies semicircle angle
AO3 (problem-solving)	5–15%	Multi-step problems combining tangent length, chord bisection and intersection of loci