Circles and Chords

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Circles and Chords — Quick Facts

Key Definitions:

• A circle is the set of all points equidistant from a fixed point called the centre
• A radius (r) is the distance from the centre to any point on the circle (r = diameter/2)
• A diameter (d) passes through the centre, connecting two points on the circle; $d = 2r$
• A chord is a line segment joining two points on the circle
• A secant is a line that cuts the circle at two points and extends beyond
• A tangent touches the circle at exactly one point and is perpendicular to the radius

Essential Formulas:

• Circumference: $C = 2\pi r = \pi d$
• Area: $A = \pi r^2$
• Arc length: $s = r\theta$ (where $\theta$ is in radians)
• Area of sector: $A = \frac{1}{2}r^2\theta$

Chord Properties:

• A perpendicular bisector of a chord passes through the centre
• Equal chords are equidistant from the centre
• The angle subtended by a chord at the centre is twice that subtended at the circumference

⚡ JAMB Exam Tip: If a question gives the radius and asks for circumference, use $C = 2\pi r$. Many candidates forget to multiply by 2 and only use $\pi r$.

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🟡 Standard — Regular Study (2d–2mo)

For students who want genuine understanding.

Circles and Chords — Study Guide

Equation of a Circle:

For a circle with centre $(h, k)$ and radius $r$: $$(x - h)^2 + (y - k)^2 = r^2$$

For a circle with centre at the origin $(0, 0)$: $$x^2 + y^2 = r^2$$

General form: $x^2 + y^2 + 2gx + 2fy + c = 0$, where centre is $(-g, -f)$ and radius $r = \sqrt{g^2 + f^2 - c}$

Example: Find the centre and radius of $x^2 + y^2 - 6x + 4y + 9 = 0$

Complete the square:

• $(x^2 - 6x) + (y^2 + 4y) = -9$
• $(x^2 - 6x + 9) + (y^2 + 4y + 4) = -9 + 9 + 4$
• $(x - 3)^2 + (y + 2)^2 = 4$

Centre = $(3, -2)$, radius = $\sqrt{4} = 2$

Perpendicular from Centre to Chord:

If a chord has length $c$ and the perpendicular distance from the centre to the chord is $d$, then: $$r^2 = d^2 + \left(\frac{c}{2}\right)^2$$

Example: Find the length of a chord of a circle of radius 5 cm at a distance of 3 cm from the centre.

Using $r^2 = d^2 + (c/2)^2$:

• $25 = 9 + (c/2)^2$
• $(c/2)^2 = 16$
• $c/2 = 4$, so $c = 8$ cm

Angles in Circles:

Position	Angle Subtended
At the centre	$2 \times$ angle at circumference
In the same segment	Equal
By diameter	$90°$ (semicircle)

Example Problem: In a circle with centre O, chord AB = 8 cm and the perpendicular distance from O to AB is 3 cm. Find the radius.

Solution: $r^2 = 3^2 + 4^2 = 9 + 16 = 25$, so $r = 5$ cm.

⚡ Common Student Mistake: Confusing when to use radians vs degrees in circle problems. JAMB typically uses degrees for angles, and radians only appear in arc length/sector area formulas where $s = r\theta$ or $A = \frac{1}{2}r^2\theta$.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Circles and Chords — Comprehensive Notes

Derivation: Perpendicular from Centre Bisects a Chord

Given chord AB with centre O. Drop perpendicular OM to chord AB.

In triangles OMA and OMB:

• $OA = OB$ (radii)
• $OM = OM$ (common)
• $\angle OMA = \angle OMB = 90°$

By RHS (Right angle-Hypotenuse-Side): $\triangle OMA \cong \triangle OMB$

Therefore, $AM = MB$ — the perpendicular from the centre bisects the chord.

Length of a Chord Formula:

For a circle of radius $r$ with a chord at perpendicular distance $d$ from centre: $$c = 2\sqrt{r^2 - d^2}$$

Alternate form: If the chord subtends an angle $\theta$ at the centre: $$c = 2r\sin\left(\frac{\theta}{2}\right)$$

Arc Length Derivation:

An arc of angle $\theta$ (in degrees) has length: $$s = \frac{\theta}{360} \times 2\pi r = \frac{\pi r \theta}{180}$$

In radians: $s = r\theta$

Sector Area Derivation: $$A = \frac{\theta}{360} \times \pi r^2 = \frac{1}{2}r^2\theta \text{ (in radians)}$$

Segment Area (Minor Segment): Area of segment = Area of sector − Area of triangle formed by two radii and the chord

$$A_{\text{segment}} = \frac{1}{2}r^2\theta - \frac{1}{2}r^2\sin\theta = \frac{1}{2}r^2(\theta - \sin\theta)$$

Intersecting Chords Theorem: If two chords AB and CD intersect at point P inside the circle: $$PA \times PB = PC \times PD$$

Example: In a circle, two chords AB and CD intersect at P. If PA = 3 cm, PB = 8 cm, and PC = 4 cm, find PD.

By intersecting chords: $3 \times 8 = 4 \times PD \Rightarrow PD = \frac{24}{4} = 6$ cm

Tangents and Secants:

If a tangent touches at point T and a secant passes through T to intersect the circle at P and Q: $$( \text{tangent} )^2 = (\text{external segment}) \times (\text{whole secant})$$ $$(PT)^2 = PA \times PQ$$

JAMB Previous Year Patterns (2015-2024):

• Finding centre and radius from general equation (2015, 2018, 2022)
• Arc length and sector area (2016, 2020, 2023)
• Chord length with given perpendicular distance (2017, 2019, 2021)
• Intersecting chords theorem (2019, 2024)

⚡ Exam Strategy: When asked for “length of chord,” draw the radius to one endpoint, the perpendicular from centre to chord, and use the right triangle. Half the chord is one leg, the perpendicular distance is the other leg, and the radius is the hypotenuse.

Grade Booster:

• Always draw a diagram for circle problems
• Label known values clearly
• Use Pythagoras theorem when you see a radius, perpendicular distance, and half-chord forming a right triangle
• Remember: $\sin(180° - \theta) = \sin\theta$, useful for angles in the same segment