Math 9 · Circle geometry
Four facts, and Pythagoras
Grade 9 circle geometry is four properties, and nearly every question is one of them plus a right triangle or a triangle with two equal sides. The skill is spotting which property the diagram is quietly handing you.
- 1. The words, first
- 2. A tangent meets the radius at 90°
- 3. The perpendicular from the centre bisects a chord
- 4. Inscribed and central angles
- 5. What costs marks
The words, first
The idea: Most of these name a line or an angle by where it sits: at the centre, on the circle, or just touching it.
| Word | What it means |
|---|---|
| Chord | A segment joining two points on the circle. |
| Radius, diameter | A segment from the centre, O, to the circle; and a chord through the centre, two radii long. |
| Arc, semicircle | Part of the circle between two points; and half a circle, cut off by a diameter. |
| Tangent | A line that touches the circle at exactly one point, the point of tangency, without crossing into it. |
| Perpendicular (⊥) | Meeting at 90°. |
| Bisect | Cut into two equal parts. |
| ∠ | “Angle.” ∠OPT is the angle at P, between PO and PT: the middle letter is the vertex. |
| Central angle | An angle with its vertex at the centre and two radii for arms. |
| Inscribed angle | An angle with its vertex on the circle and two chords for arms. |
| Standing on an arc | An angle stands on the arc between the ends of its arms. Below, ∠AOB and ∠ACB both stand on arc AB. |
| Hypotenuse | The longest side of a right triangle, opposite the right angle. Pythagoras: a² + b² = c², with c the hypotenuse. |
A tangent meets the radius at 90°
The idea: Wherever a tangent touches, the radius drawn to that point is perpendicular to it. That makes a right triangle, and a right triangle means Pythagoras.
Why it has to be true. Every point on the tangent except P is outside the circle, so OP is the shortest route from the centre to the line — and the shortest route from a point to a line is always the perpendicular one.
Worked example. PT is tangent to a circle with centre O at the point P. The radius is 5 cm and OT = 13 cm. How long is PT?
- OP is a radius to the point of tangency, so ∠OPT = 90°.
- Find the hypotenuse: OT, opposite the right angle. So OP² + PT² = OT².
- Substitute: 5² + PT² = 13², so PT² = 169 − 25 = 144, and PT = 12 cm.
Angles too. If ∠POT = 67°, then ∠PTO = 180° − 90° − 67° = 23°. The right angle is free information, and the question will not always point it out.
The trap. The property is about the radius to the point of tangency. A radius drawn anywhere else makes no promise.
The perpendicular from the centre bisects a chord
The idea: A perpendicular dropped from the centre to a chord lands exactly in the middle of it. Draw a radius to one end of the chord and there is a right triangle.
Why. OA and OB are both radii, so triangle OAB has two equal sides, and the perpendicular from O splits it into two identical right triangles. So AM = MB.
Worked example. A circle has a radius of 10 cm and a chord AB of 16 cm. How far is the chord from the centre?
- Halve the chord: OM bisects AB, so AM = 8 cm.
- Find the hypotenuse: the radius OA, 10 cm, opposite the right angle at M.
- Pythagoras: OM² = 10² − 8² = 100 − 64 = 36, so OM = 6 cm.
Backwards. A chord is 5 cm from the centre of a circle of radius 13 cm. Half the chord is √(13² − 5²) = √144 = 12 cm, so the chord is 24 cm. Doubling at the end is the step people forget.
It works in reverse too. The perpendicular bisector of any chord passes through the centre, so two chords are enough to find the centre of a broken plate.
Inscribed and central angles
The idea: An inscribed angle is half the central angle standing on the same arc. So inscribed angles on the same arc are equal, and any angle in a semicircle is 90°.
Half the central angle. In the right-hand circle of the figure above, ∠AOB = 100°, so ∠ACB, on the same arc, is 100° ÷ 2 = 50°. The other way: an inscribed angle of 35° means a central angle of 70°.
Same arc, same angle. Move C anywhere else on the big arc and the angle stays 50°, because it is always half of the same 100°.
An angle in a semicircle is 90°. If AB is a diameter, the central angle is a straight line, 180°, so an inscribed angle standing on it is 90°.
Worked example. AB is a diameter, and C is on the circle with AC = 6 cm and BC = 8 cm. Find the radius, and ∠CAB if ∠CBA = 37°.
- Spot the property: ∠ACB stands on a diameter, so it is 90°.
- Pythagoras: AB² = 6² + 8² = 100, so AB = 10 cm and the radius is 5 cm.
- Angles in a triangle add to 180°: ∠CAB = 180° − 90° − 37° = 53°.
The trap. Both angles must stand on the same arc. If C were on the small arc between A and B, the angle there would stand on the other arc and would not be 50°.
What costs marks
The idea: Halving the wrong way, and assuming a right angle that nothing gave you.
- Doubling instead of halving, or the reverse. The central angle is always the bigger one.
- The wrong hypotenuse. In the tangent triangle it is OT, from the centre to the outside point.
- Forgetting to double the half-chord at the end.
- Assuming a line bisects a chord when nothing says it comes from the centre at 90°.
- Assuming 90° with no diameter or tangent to justify it.