Question

Secants and Tangents on Circle: End-of-Topic Test
Instructions
📢 Time: 40 minutes
Instructions:
● Answer ALL questions.
● Write your answers in the spaces provided.
● Show all steps in your calculations.
● Diagrams are NOT drawn to scale unless otherwise stated.
● Equipmenu may use a scientific calculator, ruler, and compass.t: Yo
Section A: Vocabulary & Concepts
Match the term with the correct definition or visual representation. Write the letter of the correct
answer in the table.
Answer Term Definition / Property
1. Secant Line A. A line in the plane of a circle that
intersects the circle in exactly one point.
2. Tangent Line B. An angle whose vertex is on a circle and
whose sides contain chords of the circle.
3. Point of Tangency C. A line that intersects a circle in two points.
4. Chord D. The specific point where a tangent line
touches the circle.
5. Inscribed Angle E. A segment whose endpoints are on a
circle.
6. Complete the theorem: If a line is tangent to a circle, then it is __________ to the radius drawn to
the point of tangency.
Section B: Tangent Properties
7. Line AB is tangent to Circle O at point T. If the radius of the circle
is 6 cm and the length from the center O to point B is 10 cm, find
the length of the tangent segment TB.
8. Two tangent segments, PA and PB, are drawn to a circle from an
external point P. If PA = 3x + 4 and PB = 5x - 8, find the value of x
and the length of PA.
Section C: Angles formed by Secants and Tangents
For questions 9–11, find the value of the variable representing the angle or arc measure. Assume
lines that appear to be tangents are tangents.
9. Find angle x (intersection
inside circle).
10. Find angle y (intersection
outside circle).
11. Find angle z (tangent and
chord).
Section D: Segment Lengths
Apply the appropriate segment theorems (Chord-Chord, Secant-Secant, or Secant-Tangent) to
solve for x.
12. Two chords intersect inside a circle. The
segments of one chord are 4 and 9. The
segments of the other chord are x and 6. Find
x.
13. A secant and a tangent intersect outside
a circle. The tangent segment is length 10.
The external secant segment is x and the
internal secant segment is 15 (so the whole
secant is x+15).
Section E: Multi-Step Application
Context: In the diagram, AD is the diameter of circle O
extended to point P. PC is tangent to the circle at C. The
measure of angle APC is 30°.
14. Determine the measure of angle OCP. Give a reason.
15. Calculate the measure of central angle AOC.
16. Using your answer from Q15, find the measure of Arc AC.
17. Find the measure of inscribed angle ADC.
Section F: Extended Response
18. Three satellites are positioned in orbit around Earth. Satellite A can see Satellite B and
Satellite C. The lines of sight from A to B and A to C are tangent to the Earth's surface.
Explain why the distance from Satellite A to the horizon point of Satellite B must be equal to the
distance from Satellite A to the horizon point of Satellite C. Use specific geometric theorems in
your explanation.
19. A circular arch bridge has a span (chord) of 48 meters and a maximum height (sagitta) of 12
meters above the chord.
Set up an equation using the intersecting chords theorem to find the diameter of the circle that
contains this arch. (You do not need to solve the final value, just show the setup).

Answer Key
Section A: Vocabulary & Concepts
1: C, 2: A, 3: D, 4: E, 5: B
Answer:
perpendicular
Section B: Tangent Properties
Answer:
Use Pythagorean theorem: 6² + x² = 10². 36 + x² = 100. x² = 64. x = 8 cm.
Answer:
Tangents from external point are congruent. 3x + 4 = 5x - 8. 12 = 2x, x = 6. PA = 3(6) + 4 = 22.
Section C: Angles formed by Secants and Tangents
Answer:
x = (80 + 40) / 2 = 60 degrees
Answer:
y = (100 - 30) / 2 = 35 degrees
Answer:
z = 140 / 2 = 70 degrees
Section D: Segment Lengths
12: 49 = 6x -> 36 = 6x -> x=6. || 13: 10^2 = x(x+15) -> 100 = x^2 + 15x -> x^2 + 15x - 100 = 0 ->
(x+20)(x-5)=0 -> x=5 (length must be positive).
Section E: Multi-Step Application
Answer:
90 degrees. Tangent-radius theorem.
Answer:
Triangle OCP is a right triangle. 180 - 90 - 30 = 60 degrees.
Answer:
Arc AC = central angle AOC = 60 degrees.
Answer:
Angle ADC intercepts arc AC. 60 / 2 = 30 degrees.
Section F: Extended Response
Answer:
This relies on the Two-Tangent Theorem (or Hat Theorem). If two tangent segments are drawn to
a circle from the same external point, then they are congruent. Here, Satellite A is the external
point, and the lines of sight are the tangent segments. Therefore, the distances must be equal.
Answer:
Let diameter = d. The perpendicular bisector of the chord passes through the center. Half chord =
24. Sagitta = 12. The other part of the diameter segment is (d - 12). Theorem: segment * segment
= segment * segment. 24 * 24 = 12 * (d - 12).

Answer 1

Here are the answers with steps shown.

Section A: Vocabulary & Concepts
1. Secant Line — C (A line that intersects a circle in two points.)
2. Tangent Line — A (A line in the plane of a circle that intersects the circle in exactly one point.)
3. Point of Tangency — D (The specific point where a tangent line touches the circle.)
4. Chord — E (A segment whose endpoints are on a circle.)
5. Inscribed Angle — B (An angle whose vertex is on a circle and whose sides contain chords of the circle.)
6. Complete the theorem: If a line is tangent to a circle, then it is perpendicular to the radius drawn to the point of tangency.
Answer: perpendicular

Section B: Tangent Properties
7. Given: radius OT = 6 cm, OB = 10 cm, TB is tangent at T. Triangle OTB is right at T (OT ⟂ TB). Use Pythagorean theorem:
OT^2 + TB^2 = OB^2
6^2 + TB^2 = 10^2
36 + TB^2 = 100
TB^2 = 64
TB = 8 cm.

8. Tangents from an external point are congruent: PA = PB.
Given PA = 3x + 4, PB = 5x − 8. Set equal:
3x + 4 = 5x − 8
4 + 8 = 5x − 3x
12 = 2x
x = 6
PA = 3(6) + 4 = 18 + 4 = 22.

Section C: Angles formed by Secants and Tangents
(Using the standard formulas: angle formed by two chords intersecting inside = 1/2(sum of intercepted arcs); angle formed outside by two secants/tangents = 1/2(difference of intercepted arcs); angle formed by a tangent and chord = 1/2(intercepted arc).)

9. x = (80 + 40) / 2 = 60 degrees.

10. y = (100 − 30) / 2 = 35 degrees.

11. z = 140 / 2 = 70 degrees.

Section D: Segment Lengths
12. Two chords intersect inside: product of segments of one chord = product of segments of the other.
Given segments 4 and 9, and x and 6:
4 · 9 = x · 6
36 = 6x
x = 6.

13. Secant-tangent theorem: (tangent length)^2 = (external secant segment) · (whole secant).
Given tangent = 10, external secant = x, internal part = 15 (whole = x + 15):
10^2 = x(x + 15)
100 = x^2 + 15x
x^2 + 15x − 100 = 0
Factor: (x + 20)(x − 5) = 0
Positive solution: x = 5.

Section E: Multi-Step Application
Context: AD is a diameter; AP is that diameter line extended to P. PC is tangent at C. ∠APC = 30°.

14. ∠OCP = 90°. Reason: radius OC is perpendicular to tangent PC (tangent-radius theorem).

15. Find central angle AOC.
O lies on line AP because AP is the diameter line. So ∠APC = ∠OPC (O on line AP). In triangle OPC: ∠OCP = 90° and ∠OPC = 30°, so
∠POC = 180° − 90° − 30° = 60°.
Because A, O, P are collinear, ∠AOC = ∠POC = 60°. So central angle AOC = 60°.

16. Arc AC = measure of central angle AOC = 60°.

17. Inscribed angle ADC intercepts arc AC. Inscribed angle = 1/2 intercepted arc:
∠ADC = 1/2 · arc AC = 1/2 · 60° = 30°.

Section F: Extended Response
18. Satellites A, B, C with lines of sight from A to horizon points on Earth being tangents.
Explanation:
- Let O be Earth’s center and let the horizon points be H_B and H_C where the tangents from A touch the Earth.
- OB and OC are radii to the tangent points, so OB = OC.
- Radii are perpendicular to tangents at the points of tangency: OH_B ⟂ AH_B and OH_C ⟂ AH_C, so triangles OH_BA and OH_CA are right triangles.
- OA is common to both triangles.
- So each triangle has: a right angle, a shared side OA, and a radius side OB = OC. Thus the two right triangles are congruent (RHS), so AH_B = AH_C.
- Alternatively: by the Two-Tangent Theorem, tangent segments from the same external point are congruent. Therefore the distances from Satellite A to the two horizon points are equal.

19. Circular arch: chord length = 48, half-chord = 24, sagitta (height above chord to arc) = 12. Let diameter = d.
Using the intersecting-chords (or segment) setup used in the answer key:
24 · 24 = 12 · (d − 12)
(That is, 24^2 = 12 (d − 12).)
(You can also derive an equivalent equation by right-triangle geometry using radius r = d/2:
r^2 = 24^2 + (r − 12)^2, which simplifies to the same equation after algebra.)

End.