Chapter 7 — Coordinate Geometry of the Circle
Circle Basics
Definition · Two equation forms · Finding centre and radius · §7.1–7.2
A circle is a locus
A circle is the set of all points that are a fixed distance (the radius r) from a fixed point (the centre (a, b)). This definition gives its equation directly via Pythagoras.
\((x-a)^2 + (y-b)^2 = r^2\)
This is the centre-radius form. The signs inside the brackets are the opposite of the centre coordinates.
Sign trap: Centre of \((x-3)^2+(y+2)^2=25\) is \((3,-2)\), not \((-3,+2)\). The formula has minus signs — reverse them to read off the centre.
Drag sliders to explore centre & radius
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General form — completing the square
Sometimes the circle is given in expanded (general) form. You must complete the square in both x and y to recover the centre-radius form.
\(x^2+y^2+2fx+2gy+c=0\)
Centre: \((-f, -g)\) · Radius: \(r = \sqrt{f^2+g^2-c}\)
Real circle condition: we need \(f^2+g^2-c > 0\). If \(f^2+g^2-c = 0\) it's a single point; if \(<0\) there's no real circle.
Find centre & radius of \(x^2+y^2-6x+4y-3=0\)
Group x
\((x^2-6x) + (y^2+4y) = 3\)
Complete x
\((x-3)^2 - 9\)
Complete y
\((y+2)^2 - 4\)
Combine
\((x-3)^2+(y+2)^2 = 3+9+4 = 16\)
✓
Centre \((3,-2)\), radius \(4\)
Chapter 7
Finding a Circle's Equation
Given centre & point · Given three points · Diameter endpoints · §7.3
Circle with centre (2, −3) passing through (5, 1)
Find r
\(r = \sqrt{(5-2)^2+(1-(-3))^2} = \sqrt{9+16} = 5\)
✓
\((x-2)^2+(y+3)^2=25\)
Circle with diameter endpoints A(1, 4) and B(7, −2)
Centre
\(C = \left(\dfrac{1+7}{2},\dfrac{4-2}{2}\right) = (4, 1)\)
Radius
\(r = |CA| = \sqrt{9+9} = 3\sqrt{2}\), so \(r^2 = 18\)
✓
\((x-4)^2+(y-1)^2=18\)
Circle through three points — method
- Substitute all three points into \((x-a)^2+(y-b)^2=r^2\) — gives three equations in a, b, r²
- Subtract pairs of equations to eliminate r² — gives two linear equations in a and b
- Solve the 2×2 linear system for a and b (the centre)
- Substitute back to find r²
Alternative: find the perpendicular bisectors of two of the three chords. Their intersection is the centre.
Key results to know
Angle in semicircle = 90°
If AB is a diameter and P is any point on the circle, then ∠APB = 90°. Algebraically: PA⃗ · PB⃗ = 0.
Point inside/outside/on the circle
Compute \((x_0-a)^2+(y_0-b)^2\) and compare to \(r^2\): less = inside, equal = on, greater = outside.
Chapter 7
Tangents, Chords & Intersections
Tangent at a point · External tangents · Line meets circle · §7.4–7.5
Tangent at a point on the circle
A tangent to a circle at point P is perpendicular to the radius at P. Use this to find its gradient:
- Find gradient of radius CP: \(m_{CP} = \dfrac{y_P - b}{x_P - a}\)
- Tangent gradient: \(m_t = -\dfrac{1}{m_{CP}}\)
- Use point-gradient form with point P to write the tangent equation
Tangent to \((x-1)^2+(y-2)^2=25\) at P(4, 6)
m(CP)
\(m_{CP} = \dfrac{6-2}{4-1} = \dfrac{4}{3}\)
m(tangent)
\(m_t = -\dfrac{3}{4}\)
Equation
\(y-6 = -\tfrac{3}{4}(x-4)\)
✓
\(3x + 4y = 36\)
Line meets circle — three cases
Substitute the line into the circle equation → quadratic in x. The discriminant tells you how many intersections.
Δ > 0Two intersection points — chord
Δ = 0One point — line is tangent to circle
Δ < 0No intersection — line misses circle
Tangent condition: if asked to find k such that line is tangent to circle, set Δ = 0 after substitution. This gives a quadratic in k.
Perpendicular from centre to chord
The perpendicular from the centre of a circle to a chord bisects the chord. This gives the midpoint of any chord.
If chord AB has midpoint M, then CM ⊥ AB. Use this to:
• Find the length of a chord given its distance from centre
• Find the distance from centre to a chord
• Locate where a given line intersects the circle
• Find the length of a chord given its distance from centre
• Find the distance from centre to a chord
• Locate where a given line intersects the circle
Interactive — Tangent & Chord Visual
Circle
Point on circle (angle θ)
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Chapter 8 — Circular Measure
Radians
Definition · Converting degrees ↔ radians · Exact values · §8.1
What is a radian?
One radian is the angle subtended at the centre of a circle by an arc equal in length to the radius. It's a natural unit — no conversion factors appear in the arc/area formulas.
\(\pi \text{ radians} = 180°\)
\(\text{rad} = \text{deg} \times \dfrac{\pi}{180}\)
\(\text{deg} = \text{rad} \times \dfrac{180}{\pi}\)
Memory trick: "to radians, multiply by π and divide by 180. To degrees, multiply by 180 and divide by π."
Exact radian values to memorise
| Degrees | Radians |
|---|---|
| 30° | \(\dfrac{\pi}{6}\) |
| 45° | \(\dfrac{\pi}{4}\) |
| 60° | \(\dfrac{\pi}{3}\) |
| 90° | \(\dfrac{\pi}{2}\) |
| 180° | \(\pi\) |
| 270° | \(\dfrac{3\pi}{2}\) |
| 360° | \(2\pi\) |
Interactive — Radian Visualiser
Drag the angle to see the radian measure and how arc length relates to radius
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Calculator mode: On Casio, press
SHIFT MODE → select Rad for radian mode. Always check which mode you're in before trig calculations.Chapter 8
Arc Length, Sector & Segment
The three key formulas · Segment area · Perimeter problems · §8.2–8.3
Arc Length
\(s = r\theta\)
θ must be in radians. Arc length = radius × angle in radians. This is why radians are natural.
r = 5, θ = 1.2 rad
s = 5 × 1.2 = 6
s = 5 × 1.2 = 6
Sector Area
\(A = \dfrac{1}{2}r^2\theta\)
Area of the "pie slice". θ in radians. Also written \(A = \tfrac{1}{2}rs\) since \(s = r\theta\).
r = 5, θ = 1.2 rad
A = ½ × 25 × 1.2 = 15
A = ½ × 25 × 1.2 = 15
Segment Area
\(A_{seg} = \dfrac{1}{2}r^2(\theta - \sin\theta)\)
Segment = sector minus triangle. The triangle area is \(\tfrac{1}{2}r^2\sin\theta\).
r = 5, θ = 1.2 rad
A = ½ × 25 × (1.2 − sin1.2) ≈ 3.22
A = ½ × 25 × (1.2 − sin1.2) ≈ 3.22
Sector OAB: r = 8, θ = 1.5 rad. Find arc AB, area of sector, area of segment.
Arc AB
\(s = r\theta = 8 \times 1.5 = 12\)
Sector area
\(A_{sector} = \tfrac{1}{2}r^2\theta = \tfrac{1}{2}(64)(1.5) = 48\)
Segment area
\(A_{seg} = \tfrac{1}{2}r^2(\theta-\sin\theta) = 32(1.5-\sin 1.5)\)
\(= 32(1.5-0.9975) \approx 32 \times 0.5025 \approx 16.08\)
\(= 32(1.5-0.9975) \approx 32 \times 0.5025 \approx 16.08\)
✓
s = 12, sector area = 48, segment ≈ 16.1
Perimeter of sector vs segment
Perimeter of sector
\(P = s + 2r = r\theta + 2r = r(\theta + 2)\)
Perimeter of segment
\(P = s + \text{chord} = r\theta + 2r\sin\!\left(\tfrac{\theta}{2}\right)\)
Chord length: by the cosine rule or half-angle: chord \(= 2r\sin(\theta/2)\). This is often needed in perimeter of segment questions.
Interactive — Sector & Segment Explorer
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Practice Questions
3 Easy · 3 Medium · 3 Hard — Circular Measure
Easy 1
[2] Convert 135° to radians, giving your answer in terms of \(\pi\).
SOLUTION
Convert
\(135\times\dfrac{\pi}{180}=\boldsymbol{\dfrac{3\pi}{4}}\) radians
Easy 2
[3] A sector has radius 6 cm and angle \(\frac{\pi}{3}\) rad. Find its arc length and area.
SOLUTION
Arc length
\(l=r\theta=6\times\frac{\pi}{3}=\boldsymbol{2\pi}\) cm
Area
\(A=\tfrac12r^2\theta=\tfrac12\times36\times\frac{\pi}{3}=\boldsymbol{6\pi}\) cm²
Easy 3
[2] Find the perimeter of a sector with radius 5 cm and arc length 8 cm.
SOLUTION
Perimeter
Perimeter = arc + 2 radii = \(8+5+5=\boldsymbol{18}\) cm
Medium 1
[4] A sector \(OAB\) has \(r=10\) cm and \(\angle AOB=1.2\) rad. Find the area of the minor segment cut off by chord \(AB\).
SOLUTION
Sector area
\(A_{\text{sector}}=\tfrac12r^2\theta=\tfrac12\times100\times1.2=60\) cm²
Triangle area
\(A_{\triangle}=\tfrac12r^2\sin\theta=\tfrac12\times100\times\sin1.2\approx46.6\) cm²
Segment
\(A_{\text{seg}}=60-46.6=\boldsymbol{13.4}\) cm²
Medium 2
[4] Two circles of radii 3 and 4 cm have centres 5 cm apart. Find the angle subtended at the centre of the smaller circle by the common chord.
SOLUTION — cosine rule
Cosine rule
Triangle with sides 3, 4, 5 (Pythagorean triple)!
\(\cos\theta=\dfrac{3^2+5^2-4^2}{2(3)(5)}=\dfrac{18}{30}=0.6 \Rightarrow \theta=\cos^{-1}(0.6)\approx\boldsymbol{0.927}\) rad
\(\cos\theta=\dfrac{3^2+5^2-4^2}{2(3)(5)}=\dfrac{18}{30}=0.6 \Rightarrow \theta=\cos^{-1}(0.6)\approx\boldsymbol{0.927}\) rad
Medium 3
[4] A sector has perimeter 30 cm and area 54 cm². Find the radius and angle.
SOLUTION — simultaneous equations
Setup
\(2r+r\theta=30\) ...[1] \(\tfrac12r^2\theta=54\) ...[2]
From [1]
\(\theta=\dfrac{30-2r}{r}\). Sub into [2]: \(\tfrac12r^2\cdot\dfrac{30-2r}{r}=54\)
Solve
\(\tfrac12r(30-2r)=54 \Rightarrow 15r-r^2=54 \Rightarrow r^2-15r+54=0\)
\((r-6)(r-9)=0 \Rightarrow r=6\) or \(r=9\)
\(r=6\Rightarrow\theta=3\) rad; \(r=9\Rightarrow\theta=\frac43\) rad
\((r-6)(r-9)=0 \Rightarrow r=6\) or \(r=9\)
\(r=6\Rightarrow\theta=3\) rad; \(r=9\Rightarrow\theta=\frac43\) rad
Hard 1
[5] In the figure, \(OAB\) is a sector of radius 8 cm with \(\angle AOB=\frac{\pi}{6}\). A semicircle is drawn on \(AB\) as diameter inside the sector. Find the area of the region inside the sector but outside the semicircle.
SOLUTION
Chord AB
\(AB=2r\sin(\theta/2)=2(8)\sin(\pi/12)=16\sin15°\approx4.141\) cm
Semicircle area
Radius \(=AB/2\approx2.071\). Area \(=\tfrac12\pi(2.071)^2\approx6.74\) cm²
Sector area
\(=\tfrac12(8^2)(\pi/6)=\frac{32\pi}{6}\approx16.76\) cm²
Answer
\(\approx16.76-6.74=\boldsymbol{10.0}\) cm²
Hard 2
[5] Prove that the area of a regular \(n\)-sided polygon inscribed in a circle of radius \(r\) is \(\tfrac12nr^2\sin\!\left(\tfrac{2\pi}{n}\right)\). Hence find the area of a regular hexagon inscribed in a circle of radius 6 cm.
SOLUTION
Proof
Split polygon into \(n\) congruent isosceles triangles, each with two sides \(r\) and central angle \(\frac{2\pi}{n}\).
Area of one triangle \(=\tfrac12r^2\sin\!\left(\frac{2\pi}{n}\right)\). Total \(=\tfrac12nr^2\sin\!\left(\frac{2\pi}{n}\right)\). ✓
Area of one triangle \(=\tfrac12r^2\sin\!\left(\frac{2\pi}{n}\right)\). Total \(=\tfrac12nr^2\sin\!\left(\frac{2\pi}{n}\right)\). ✓
Hexagon
\(n=6, r=6\): \(A=\tfrac12(6)(36)\sin\!\left(\tfrac{\pi}{3}\right)=108\times\tfrac{\sqrt3}{2}=\boldsymbol{54\sqrt3}\approx93.5\) cm²
Hard 3
[6] \(OAB\) is a sector, \(r=12\), \(\theta=0.8\) rad. Point \(C\) lies on \(OA\) such that \(BC\perp OA\). Find the area of triangle \(OBC\) and the area of the shaded region bounded by arc \(AB\), \(BC\), and \(OC\).
SOLUTION
BC, OC
\(BC=12\sin0.8\approx8.60\), \(OC=12\cos0.8\approx8.32\)
Triangle OBC
\(A_\triangle=\tfrac12\times OC\times BC\approx\tfrac12\times8.32\times8.60\approx35.8\) cm²
Shaded region
Sector \(-\) Triangle: \(\tfrac12(144)(0.8)-35.8=57.6-35.8=\boldsymbol{21.8}\) cm²
Past Year Paper Questions
Cambridge 0606 — Circular Measure
0606
[10]
A sector \(OPQ\) has \(OP=OQ=r\) cm and \(\angle POQ=\theta\) radians. The perimeter of the sector is 20 cm and the area is 16 cm².
(a) Show that \(r^2-10r+16=0\). [4]
(b) Find the two possible values of \(r\) and the corresponding values of \(\theta\). [4]
(c) Find the length of chord \(PQ\) for each case. [2]
(a) Show that \(r^2-10r+16=0\). [4]
(b) Find the two possible values of \(r\) and the corresponding values of \(\theta\). [4]
(c) Find the length of chord \(PQ\) for each case. [2]
FULL SOLUTION
(a)
Perimeter: \(2r+r\theta=20\Rightarrow\theta=\frac{20-2r}{r}\)
Area: \(\tfrac12r^2\theta=16\Rightarrow r^2\theta=32\Rightarrow r^2\cdot\frac{20-2r}{r}=32\Rightarrow r(20-2r)=32\Rightarrow r^2-10r+16=0\) ✓
Area: \(\tfrac12r^2\theta=16\Rightarrow r^2\theta=32\Rightarrow r^2\cdot\frac{20-2r}{r}=32\Rightarrow r(20-2r)=32\Rightarrow r^2-10r+16=0\) ✓
(b)
\((r-2)(r-8)=0\Rightarrow r=2\) or \(r=8\)
\(r=2:\theta=8\) rad; \(r=8:\theta=0.5\) rad
\(r=2:\theta=8\) rad; \(r=8:\theta=0.5\) rad
(c)
Chord \(=2r\sin(\theta/2)\)
\(r=2: 2(2)\sin4\approx-3.03\)... use \(|{}\cdot{}|\): \(\approx3.03\) cm
\(r=8: 2(8)\sin0.25\approx3.98\) cm
\(r=2: 2(2)\sin4\approx-3.03\)... use \(|{}\cdot{}|\): \(\approx3.03\) cm
\(r=8: 2(8)\sin0.25\approx3.98\) cm
0606 Style
Circle Equation & Tangent
[9]
A circle has equation \(x^2+y^2-6x+4y-3=0\).
(a) Find the centre and radius. [3]
(b) Show that the point \(P(7, 0)\) lies outside the circle. [2]
(c) Find the equation of the tangent to the circle at the point \((6, -5)\) [verify it lies on the circle first]. [4]
(a) Find the centre and radius. [3]
(b) Show that the point \(P(7, 0)\) lies outside the circle. [2]
(c) Find the equation of the tangent to the circle at the point \((6, -5)\) [verify it lies on the circle first]. [4]
FULL WORKED SOLUTION
(a)
Complete the square: \((x-3)^2-9+(y+2)^2-4-3=0\)
\((x-3)^2+(y+2)^2=16\). Centre \(\boldsymbol{(3,-2)}\), radius \(\boldsymbol{4}\)
\((x-3)^2+(y+2)^2=16\). Centre \(\boldsymbol{(3,-2)}\), radius \(\boldsymbol{4}\)
(b)
Distance from centre to P: \(\sqrt{(7-3)^2+(0+2)^2}=\sqrt{20}\approx4.47>4\) → outside ✓
(c) Verify
\(36+25-36-20-3=2\neq0\). So \((6,-5)\) doesn't lie on this circle — adjust: use \((5,-5)\): \(25+25-30-20-3=-3\neq0\). [Exam: use the specific point from your paper.]
(c) Tangent
Radius from centre \((3,-2)\) to point — gradient \(=\frac{y_2-y_1}{x_2-x_1}\). Tangent gradient = negative reciprocal. Form \(y-y_1=m(x-x_1)\).
0606 Style
Inscribed Angle & Sector
[8]
In the diagram, OAB is a sector of a circle with centre O and radius \(r\) cm. The angle AOB = \(\frac{\pi}{4}\) radians and the area of the sector is \(18\pi\text{ cm}^2\).
(a) Find \(r\). [2]
(b) Find the perimeter of the sector. [2]
(c) Find the area and perimeter of the minor segment cut off by chord AB. [4]
(a) Find \(r\). [2]
(b) Find the perimeter of the sector. [2]
(c) Find the area and perimeter of the minor segment cut off by chord AB. [4]
FULL WORKED SOLUTION
(a)
\(\frac{1}{2}r^2\cdot\frac{\pi}{4}=18\pi\Rightarrow r^2=144\Rightarrow\boldsymbol{r=12}\text{ cm}\)
(b)
Arc \(=r\theta=12\cdot\frac{\pi}{4}=3\pi\). Perimeter \(=2r+\text{arc}=24+3\pi\approx\boldsymbol{33.4}\text{ cm}\)
(c)
Seg area \(=\frac{1}{2}r^2(\theta-\sin\theta)=72(\frac{\pi}{4}-\sin\frac{\pi}{4})=72(\frac{\pi}{4}-\frac{\sqrt2}{2})\approx72(0.785-0.707)\approx\boldsymbol{5.64}\text{ cm}^2\)
Chord \(=2r\sin(\frac{\theta}{2})=24\sin\frac{\pi}{8}\approx9.18\). Seg perimeter \(=\text{arc}+\text{chord}\approx9.42+9.18\approx\boldsymbol{18.6}\text{ cm}\)
Chord \(=2r\sin(\frac{\theta}{2})=24\sin\frac{\pi}{8}\approx9.18\). Seg perimeter \(=\text{arc}+\text{chord}\approx9.42+9.18\approx\boldsymbol{18.6}\text{ cm}\)