ACT Math study guide · Chapter 28
Circle Theorems
An inscribed angle, with its vertex on the circle, is always half the central angle that intercepts the same arc, shown below. A direct consequence is that any angle inscribed in a semicircle, with a diameter as one side, always measures 90°, and any two inscribed angles that intercept the same arc are equal to each other.
A tangent line touches a circle at exactly one point. A tangent-chord angle, formed where a tangent meets a chord at that point of tangency, equals half the intercepted arc, the same relationship as an inscribed angle. Two tangent segments drawn from the same external point to a circle are always equal in length.
Chords, secants, and tangents that intersect inside or outside a circle follow three distinct relationships, and mixing them up is the most common error on this topic. Two chords intersecting inside a circle: the products of their segments are equal. Two secants (or a tangent and secant) meeting outside a circle: for secants, the angle formed equals half the difference of the intercepted arcs, and the products of each external part times its whole length are equal; for a tangent and secant, the tangent length squared equals the external part times the whole secant length.
Before calculating anything, identify what the figure actually shows. The table below matches each visual signal to its relationship, so recognition happens before arithmetic, not during it.
| What the figure shows | Relationship to apply |
|---|---|
| The center of the circle is involved in the angle | A central angle equals its intercepted arc |
| The angle's vertex sits on the circle itself | An inscribed angle is half its intercepted arc |
| Two chords cross inside the circle | The products of each chord's two segments are equal |
| Two secants meet at a point outside the circle | External segment × whole secant is equal for both secants |
| A tangent and a secant share the same exterior point | Tangent² = external secant × whole secant |
| A tangent meets a radius at the point of tangency | The angle between them is 90° |
Do not try to memorize these as one giant circle formula. Matching the picture to the correct relationship is the actual skill; once that match is made, the calculation itself is usually short.
When a tangent and a secant form an angle outside the circle, the angle measure is one-half the difference of the intercepted arcs, just as for two secants from an external point.
Where this appears as a model: a support beam across a circular structure. A circular window with a beam creating an inscribed angle across a 100° arc has a beam angle of 50°, half the arc it intercepts.
▸ Inscribed and central angles
Worked Example A central angle measures 96°. Find the measure of an inscribed angle that intercepts the same arc.
An inscribed angle is half the central angle intercepting the same arc.
96° ÷ 2 = 48°
▸ Intersecting chords
Worked Example Two chords intersect inside a circle. The segments of one chord measure 6 and 10; the segments of the other measure 5 and x. Find x.
The products of the two chords' segments are equal.
6 × 10 = 5 × x
60 = 5x
x = 12
▸ Tangent and secant from an external point
Worked Example A tangent segment of length 6 and a secant are drawn from the same point outside a circle. The external part of the secant measures 4. Find the whole length of the secant.
Tangent² equals the external part times the whole secant length.
6² = 4 × x
36 = 4x
x = 9
Worked Example A tangent segment of length 12 and a secant are drawn from the same point outside a circle. The secant's external part measures 8. Find the length of the secant's internal (far) segment.
Recognize. This is a tangent-secant relationship: tangent² equals the external part times the whole secant length. The question asks for the internal segment, not the whole length, so an extra step remains after finding the whole.
Set up.
12² = 8 × (whole length)
Solve.
144 = 8 × (whole length)
whole length = 18
internal segment = 18 − 8 = 10
Check. external × whole = 8 × 18 = 144 = 12², which matches the original relationship.
Best route. Hand algebra. Straightforward once the tangent-secant relationship is recognized.
Common error. Stopping at the whole secant length (18) instead of subtracting the external part to find the requested internal segment (10).
▸ Practice
Question 1
- A. 20°
- B. 40°
- C. 80°
- D. 160°
Question 2
- F. 130°
- G. 260°
- H. 65°
- J. 50°
Question 3
- A. 45°
- B. 60°
- C. 180°
- D. 90°
Question 4
- F. 55°
- G. 35°
- H. 70°
- J. 17.5°
Question 5
- A. 50°
- B. 200°
- C. 100°
- D. 25°
Question 6
- F. 12
- G. 18
- H. 24
- J. 6
Question 7
- A. 8
- B. 4.5
- C. 7
- D. 9
Question 8
- F. 180°
- G. 40°
- H. 90°
- J. 20°
Question 9
- A. 70°
- B. 35°
- C. 65°
- D. 130°
Question 10
- F. 12
- G. 16
- H. 2
- J. 32
Question 11
- A. 37.5°
- B. 150°
- C. 300°
- D. 75°
Question 12
- F. 3.75
- G. 6.67
- H. 4
- J. 60
Question 13
- A. 168°
- B. 84°
- C. 21°
- D. 42°
Question 14
- F. 145°
- G. 55°
- H. 125°
- J. 35°
Question 15
- A. 6.4
- B. 12
- C. 10
- D. 9
Question 16
- F. 55°
- G. 220°
- H. 27.5°
- J. 110°
Question 17
- A. 4
- B. 16
- C. 6
- D. 9
Question 18
- F. 50°
- G. 100°
- H. 75°
- J. 200°
Question 19
- A. 12
- B. 3
- C. 27
- D. 24
Question 20
- F. 48°
- G. 24°
- H. 192°
- J. 96°
▸ Answers and Explanations
Show the answers and explanations
1: B. An inscribed angle is half the central angle that intercepts the same arc: 80° ÷ 2 = 40°. Choice C reports the central angle itself, forgetting to halve it.
2: H. An inscribed angle is half the arc it intercepts: 130° ÷ 2 = 65°. Choice G doubles the arc instead of halving it.
3: D. An angle inscribed in a semicircle, opposite a diameter, always measures 90°. Choice C reports the full semicircle's arc measure (180°) instead of the inscribed angle.
4: G. Two inscribed angles that intercept the same arc are always equal: the other angle also measures 35°. Choice F doubles the angle, mistaking it for the central angle that intercepts the same arc.
5: A. A tangent-chord angle equals half the intercepted arc: 100° ÷ 2 = 50°. Choice C reports the intercepted arc itself, forgetting to halve it.
6: F. The two tangent segments drawn from the same external point to a circle are always equal in length. Choice H doubles the given length instead of recognizing the two tangent segments are equal.
7: A. When two chords intersect inside a circle, the products of their segments are equal: 4 × 6 = 3 × x, so x = 24/3 = 8. Choice C adds and subtracts the segment lengths instead of setting the two products equal.
8: H. The angle formed by two intersecting chords equals half the sum of the two intercepted arcs: (70°+110°)/2 = 90°. Choice J uses the difference of the arcs instead of their sum.
9: B. The angle formed by two secants from an external point equals half the difference of the two intercepted arcs: (100°−30°)/2 = 35°. Choice A uses the sum of the arcs instead of their difference, applying the chord-chord rule instead of the secant-secant rule.
10: G. For a tangent and secant from the same external point, tangent² = external × whole: 8² = 4 × x, so x = 64/4 = 16. Choice H forgets to square the tangent length, solving 8 = 4x instead of 8² = 4x.
11: D. An inscribed angle is half the central angle that intercepts the same arc: 150° ÷ 2 = 75°. Choice B reports the central angle itself, forgetting to halve it.
12: F. For two secants from the same external point, the products of each external part and whole length are equal: 5 × 15 = x × 20, so x = 75/20 = 3.75. Choice G sets up the relationship as a ratio of the parts instead of setting the two products equal.
13: B. A central angle is twice the inscribed angle that intercepts the same arc: 42° × 2 = 84°. Choice D reports the inscribed angle itself, forgetting to double it.
14: G. The angle opposite the diameter is 90°, since it is inscribed in a semicircle. The three angles of the triangle sum to 180°: 180° − 90° − 35° = 55°. Choice F subtracts only the given 35° from 180°, forgetting that the angle opposite the diameter is 90°.
15: C. When two chords intersect inside a circle, the products of their segments are equal: 5 × 8 = 4 × x, so x = 40/4 = 10. Choice A sets up the relationship as a ratio of the parts instead of setting the two products equal.
16: J. A tangent-chord angle equals half the intercepted arc, so the arc is twice the angle: 55° × 2 = 110°. Choice H halves the angle instead of doubling it to find the arc.
17: A. For two secants from the same external point, the products of each external part and whole length are equal: 6 × 16 = x × 24, so x = 96/24 = 4. Choice D pairs the external part of one secant with the whole length of the other incorrectly, computing 6×24/16=9 instead.
18: F. The angle formed by two secants from an external point equals half the difference of the two intercepted arcs: (150°−50°)/2 = 50°. Choice G uses the sum of the arcs instead of their difference, applying the chord-chord rule instead of the secant-secant rule.
19: C. For a tangent and secant from the same external point, tangent² = external × whole: 9² = 3 × x, so x = 81/3 = 27. Choice B forgets to square the tangent length, solving 9 = 3x instead of 9² = 3x.
20: J. Each inscribed angle is half the intercepted arc: 96° ÷ 2 = 48°. The sum of the two equal angles is 48° + 48° = 96°. Choice F reports the measure of one inscribed angle alone instead of the sum of both.