0580 · 4.7Circle theorems I

Angles at the centre

0:00A circle can connect angles whose vertices are far apart. In this diagram, knowing the angle at the centre lets us calculate the angle at the top. We do not need the circle’s radius or a protractor. The shape itself provides enough information, once we recognise the right relationship. By the end, you will understand six circle theorems and how they fit together. We will calculate unknown angles and give geometrical reasons for our answers.

0:23Most importantly, we will connect these rules to equal radii, triangle angles and arcs. That makes unfamiliar diagrams much easier to interpret. We need three familiar angle facts. Angles in a triangle sum to one hundred and eighty degrees, and angles around a point sum to three hundred and sixty degrees. Angles on a straight line sum to one hundred and eighty degrees. An isosceles triangle has two equal sides and two equal base angles.

0:48The circumference is the boundary of a circle. A radius joins the centre to the circumference, and a diameter passes through the centre with both endpoints on the circle. A chord joins any two points on the circumference. An arc is part of that boundary, while a tangent touches the circle at one point. All radii of the same circle are equal. Joining the centre to two points on the circumference therefore creates an isosceles triangle.

1:12This simple fact explains much of the angle structure we are about to use. Whenever the centre is shown, look for triangles formed by two radii. Our first relationship compares an angle at the centre with an angle at the circumference. Both angles must stand on the same arc. Here, their arms lead to P and Q, and the relevant arc runs between those points below the centre. The angle at the centre is twice the angle at the circumference, when both stand on the same arc.

1:36We can see why by joining R to the centre. Let angle PRO be x degrees, and angle ORQ be y degrees. Triangle OPR is isosceles because its two radii are equal. Its angle at the centre is therefore one hundred and eighty degrees minus 2x degrees. The other isosceles triangle gives angle ROQ as one hundred and eighty degrees minus 2y degrees. Subtracting both angles from a full turn leaves angle POQ.

2:03The two amounts of one hundred and eighty degrees cancel the full turn. What remains is 2x degrees plus 2y degrees. At the circumference, angle PRQ is x degrees plus y degrees. The centre angle is therefore twice that whole angle. This diagram shows the reasoning for one arrangement. The theorem also works when the centre lies outside the triangle. The essential condition is using the same arc. Returning to our opening example, the centre angle is one hundred degrees.

2:31Dividing by two gives fifty degrees at the circumference. If the circumference angle were given instead, we would multiply it by two. Identify which vertex is the centre before deciding whether to double or halve.

Semicircles and same segments

2:43A diameter gives an especially useful case. The angle at the centre along the diameter is one hundred and eighty degrees. The circumference angle standing on that semicircle is half as large. This gives our second theorem: the angle in a semicircle is ninety degrees. The right angle is at the point on the circumference, opposite the diameter. It is not automatically at either endpoint of the diameter. Also, the side must genuinely pass through the centre.

3:07A long chord that merely looks like a diameter is not enough to justify this theorem. For our first worked example, PQ is a diameter and angle RPQ is thirty-four degrees. We need angle PQR. First, angle PRQ is ninety degrees because the angle in a semicircle is ninety degrees. That supplies the missing information needed to use the triangle angle sum. The unknown angle equals one hundred and eighty degrees minus ninety degrees minus thirty-four degrees.

3:36Subtracting ninety degrees leaves ninety degrees. Taking away thirty-four degrees then gives fifty-six degrees. Our answer needs both reasons: angle in a semicircle, followed by angles in a triangle. A chord divides the circle into two segments. Each segment is the region between that chord and one of its arcs. In this diagram, R and S lie in the same segment. The angles they form using the chord’s endpoints are equal, even though their vertices occupy different positions.

4:04The theorem is that angles in the same segment are equal. Both circumference angles are half the same centre angle, so they must match. You can imagine moving the vertex along the same arc while keeping the chord fixed. The angle stays unchanged throughout that movement. The words same segment matter. If the vertices lie on opposite sides of the chord, these angles are not generally equal. Also, the two angles must use exactly the same chord endpoints.

4:29Two angles somewhere on the same circle do not automatically have a relationship. In our second example, the reflex angle at the centre is two hundred and fifty degrees. The required circumference angle has its vertex on the short arc below the chord. Its arms stand on the long arc around the top. We must therefore use the reflex centre angle, not the smaller one. The angle at the centre is twice the angle at the circumference.

4:53So angle PRQ = 250 degrees divided by two, giving one hundred and twenty-five degrees. Angle PSQ equals angle PRQ because they are angles in the same segment. The second unknown is therefore also one hundred and twenty-five degrees. This example prevents a common mistake: always choosing the smaller centre angle. Trace the arc between the endpoints that does not contain the circumference vertex. That is the arc used by the circumference angle.

5:21Its corresponding centre angle can be reflex, meaning greater than one hundred and eighty degrees.

Cyclic quadrilaterals

5:26A cyclic quadrilateral is a quadrilateral whose four vertices lie on a circle. Cyclic describes the positions of the vertices, not the shape of the sides. It does not need to be a rectangle or have equal sides. Its special angle relationship applies to each pair of opposite angles. Opposite angles of a cyclic quadrilateral sum to one hundred and eighty degrees. We also say that these angles are supplementary.

5:49To understand why, look at the opposite vertices P and R. Both angles use the endpoints Q and S, but they stand on the two different arcs between them. Those arcs together make a complete circle. Their corresponding centre angles therefore add to three hundred and sixty degrees. Each circumference angle is half its corresponding centre angle. Their sum must consequently be one hundred and eighty degrees, which explains the cyclic quadrilateral theorem.

6:14Our third example combines circle geometry with algebra. The opposite angles are 2x + 10 degrees and 3x + 20 degrees. Because the quadrilateral is cyclic, we add those expressions and set the total equal to one hundred and eighty degrees. Collecting like terms gives 5x + 30 = 180. Subtract thirty from both sides to obtain 5x = 150. Dividing by five gives x = 30. The first angle is two times thirty plus ten, giving seventy degrees.

6:43For the opposite angle, three times thirty plus twenty gives one hundred and ten degrees. Our check is seventy degrees plus one hundred and ten degrees equals one hundred and eighty degrees. Notice that x was not itself either angle. We substituted its value into both expressions to answer the question fully. Also, the theorem applies to opposite angles, not neighbouring ones.

Tangent angle theorems

7:04A tangent gives another right angle. The angle between a tangent and the radius at the point of contact is ninety degrees. The location is essential: the radius must end exactly where the tangent touches. A radius drawn to another point does not have this perpendicular relationship. There is a useful geometric reason behind this. The perpendicular from the centre to a straight line gives the shortest distance to that line.

7:27If this distance were smaller than the radius, the line would cut through the circle. At a tangent, the shortest distance equals the radius, so the radius meets it perpendicularly. The final theorem connects a tangent, a chord and a circumference angle. The angle between a tangent and a chord is equal to the angle in the alternate segment. Here, the angle between the tangent towards T and chord PQ equals angle PRQ.

7:51This relationship is called the alternate segment theorem. To locate the matching angle, start at the chord’s two endpoints. Join them to a point on the circumference on the other side of the chord from the tangent angle. That creates the required circumference angle. Do not choose an angle simply because it looks similar or sits near the tangent. We can connect this theorem to our earlier results. Suppose the acute tangent angle is alpha degrees.

8:15The radius is perpendicular to the tangent, so angle OPQ is ninety degrees minus alpha degrees. The equal radii make angle OQP the same size. The triangle angle sum then makes angle POQ equal to two alpha degrees. The circumference angle standing on that same arc is half the centre angle. It is therefore alpha degrees, matching the tangent angle. The same theorem also applies to the other tangent ray, with its own alternate segment.

8:40Our fourth example has a tangent at P and two given triangle angles. Angle PRQ is thirty-eight degrees, while angle RPQ is sixty-seven degrees. To find angle PQR, subtract both from one hundred and eighty degrees. This gives one hundred and forty-two degrees minus sixty-seven degrees, which equals seventy-five degrees. Angle TPQ equals angle PRQ, so it is thirty-eight degrees by the alternate segment theorem. Angle OPT is ninety degrees, because the tangent is perpendicular to the radius.

9:12Subtracting thirty-eight degrees from ninety degrees gives angle OPQ as fifty-two degrees. Each answer follows from a specific relationship, not the diagram’s appearance. These are typical examination tasks: calculate unknown angles and give explanations using geometrical properties. A creditworthy answer names the theorem as well as giving the value. For the tangent angle, write that it is thirty-eight degrees by the alternate segment theorem. For the radius calculation, state that the tangent is perpendicular to the radius at the point of contact.

Exam technique and recap

9:41All the arithmetic we have used must be manageable without a calculator on Paper two. The same reasoning also applies on calculator Paper four. Our answers here are exact, so no rounding is needed. If another question requires approximation, retain intermediate values and follow its stated accuracy instructions. The most common errors come from matching the wrong parts of the diagram. Check that a supposed diameter passes through the centre and that a tangent really touches at the radius endpoint.

10:07For circumference angles, identify the chord and the relevant arc. For cyclic quadrilaterals, pair opposite vertices rather than neighbouring ones. The central idea is that equal radii create predictable triangles, while arcs connect angles across the circle. A centre angle is twice the circumference angle standing on the same arc. A semicircle therefore produces ninety degrees, and angles in the same segment are equal. Opposite cyclic angles sum to one hundred and eighty degrees.

10:33At a tangent, the radius is perpendicular at the point of contact. The alternate segment theorem links a tangent and chord angle to the matching circumference angle. Identify the structure first, calculate second, and attach the correct reason. These connections turn a collection of rules into a coherent way of understanding circle geometry.