GRADE 10 · A4 PRINT VIEW
Circles · Answer guide
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Lesson 1 · Locate the radius perpendicular to a tangent
Identify a tangent and apply the right angle at its contact point.
A line touches a circle at T, whose centre is O. What angle does OT make with that tangent?
Answer: 90°. The radius through the point of contact is perpendicular to the tangent.
OT = 5 cm is a radius, PT is tangent at T, and OP = 13 cm. Find PT.
Answer: 12 cm. Triangle OTP is right-angled at T, so PT = √(13² − 5²) = √144 = 12 cm.
A straight line meets a circle at two distinct points. What is it called?
Answer: A secant. A secant intersects the circle twice; a tangent meets it at exactly one point.
Discuss: Draw a tangent at the top of a circle. Show which radius creates the right angle and distinguish a parallel line that misses the circle.
Teaching note: A line perpendicular to a radius is the tangent only when it passes through that radius endpoint on the circle. Open responses need discussion; they are not automatically marked.
Lesson 2 · Compare tangents from one external point
Use equal tangent lengths and the congruent-triangle reason.
From external point P, tangents PA and PB touch the same circle. If PA = 4 cm, find PB.
Answer: 4 cm. Tangent segments drawn from the same external point to the same circle have equal lengths, so PB = PA = 4 cm.
From external point P, tangents PA and PB touch the same circle. If PA = 7 cm, find PB.
Answer: 7 cm. Tangent segments drawn from the same external point to the same circle have equal lengths, so PB = PA = 7 cm.
From external point P, tangents PA and PB touch the same circle. If PA = 9 cm, find PB.
Answer: 9 cm. Tangent segments drawn from the same external point to the same circle have equal lengths, so PB = PA = 9 cm.
Discuss: Sketch two different outside points. Explain why the theorem pairs two tangent lengths at each point, not all four lengths together.
Teaching note: Tangent lengths are equal from the same external point. Tangents from different external points need not have equal lengths. Open responses need discussion; they are not automatically marked.
Lesson 3 · Connect central and tangent angles
Combine two tangent right angles with an angle sum.
PA and PB are tangents from P. The non-reflex central angle AOB is 80°. Find angle APB.
Answer: 100°. In quadrilateral OAPB, contact angles at A and B are 90°. Thus APB = 360° − 90° − 90° − 80° = 100°.
PA and PB are tangents from P. The non-reflex central angle AOB is 90°. Find angle APB.
Answer: 90°. In quadrilateral OAPB, contact angles at A and B are 90°. Thus APB = 360° − 90° − 90° − 90° = 90°.
PA and PB are tangents from P. The non-reflex central angle AOB is 100°. Find angle APB.
Answer: 80°. In quadrilateral OAPB, contact angles at A and B are 90°. Thus APB = 360° − 90° − 90° − 100° = 80°.
Discuss: Choose three possible central angles between 0° and 180°. Calculate the tangent angle and describe how the two change together.
Teaching note: Do not subtract a reflex central angle from 180°. Identify the angle inside the quadrilateral. Open responses need discussion; they are not automatically marked.
Final check · Set 1
Try one set after the lessons. Use the other set after reviewing any errors.
OT = 3 cm, OP = 5 cm and PT is tangent at T. Find PT.
4 cm. PT² = OP² − OT² = 25 − 9 = 16, so PT = 4 cm.
In right triangle OTP formed by centre O, contact T and external point P, which side is the hypotenuse?
OP. The angle at T is 90°, so the side opposite it, OP, is the hypotenuse.
From external point P, tangents PA and PB touch the same circle. If PA = 15 cm, find PB.
15 cm. Tangent segments drawn from the same external point to the same circle have equal lengths, so PB = PA = 15 cm.
From external point P, tangents PA and PB touch the same circle. If PA = 18 cm, find PB.
18 cm. Tangent segments drawn from the same external point to the same circle have equal lengths, so PB = PA = 18 cm.
PA and PB are tangents from P. The non-reflex central angle AOB is 120°. Find angle APB.
60°. In quadrilateral OAPB, contact angles at A and B are 90°. Thus APB = 360° − 90° − 90° − 120° = 60°.
PA and PB are tangents from P. The non-reflex central angle AOB is 130°. Find angle APB.
50°. In quadrilateral OAPB, contact angles at A and B are 90°. Thus APB = 360° − 90° − 90° − 130° = 50°.
Final check · Set 2
Try one set after the lessons. Use the other set after reviewing any errors.
A tangent segment PT is 15 cm and radius OT is 8 cm. Find OP.
17 cm. OP² = 15² + 8² = 289, so OP = 17 cm.
A tangent segment PT is 8 cm and radius OT is 6 cm. Find OP.
10 cm. OP² = 8² + 6² = 100, so OP = 10 cm.
From external point P, tangents PA and PB touch the same circle. If PA = 20 cm, find PB.
20 cm. Tangent segments drawn from the same external point to the same circle have equal lengths, so PB = PA = 20 cm.
From external point P, tangents PA and PB touch the same circle. If PA = 23 cm, find PB.
23 cm. Tangent segments drawn from the same external point to the same circle have equal lengths, so PB = PA = 23 cm.
PA and PB are tangents from P. The non-reflex central angle AOB is 140°. Find angle APB.
40°. In quadrilateral OAPB, contact angles at A and B are 90°. Thus APB = 360° − 90° − 90° − 140° = 40°.
PA and PB are tangents from P. The non-reflex central angle AOB is 150°. Find angle APB.
30°. In quadrilateral OAPB, contact angles at A and B are 90°. Thus APB = 360° − 90° − 90° − 150° = 30°.
Original learning companion, not full textbook exercise solutions. Covers the radius–tangent relation, equal tangent lengths with RHS reasoning, and a tangent-angle consequence. It provides a concise proof explanation, not every textbook proof or exercise configuration.