Chapter 9.3 — Exercise 9.2 — Construction of Tangents
Construction of pair of tangents from an external point. This is Lesson 3 of 4 in Chapter 9: Tangents and Secants to a Circle.
How Many Tangents Fit Through One Point
Where a point sits relative to a circle decides everything about how many tangent lines can reach it. Exercise 9.2 works out exactly that count, proves a genuinely useful fact about the two tangents an outside point always gets, and constructs them directly with compass and straightedge.
Zero, One, or Two — Never Anything Else
- A point inside the circle: no tangent can be drawn through it at all. Every line through an interior point is forced to cross the circle twice, making it a secant no matter which direction it's drawn.
- A point on the circle: exactly one tangent exists — the line perpendicular to the radius at that exact point.
- A point outside the circle: exactly two tangents can be drawn, one touching each "visible side" of the circle from that point's perspective.
The Two Tangents From an External Point Are Always Equal
∠OAP = ∠OBP = 90° OA = OB (radii) OP = OP (common)
⟹ △OAP ≅ △OBP (RHS) ⟹ PA = PBThe Length of Tangents Theorem: the two right triangles share the hypotenuse OP and have equal legs OA and OB, which is exactly the RHS (right angle–hypotenuse–side) congruency condition. Once the triangles are congruent, PA=PB follows immediately as corresponding parts of congruent triangles — no separate argument needed. The same congruency also hands over a second fact almost for free: OP bisects ∠APB, since the two triangles are mirror images of each other across line OP. That bisected-angle fact turns out to matter later in this exercise, whenever a problem hands over the full angle between two tangents and expects half of it used directly.
Angle Facts, Straight From the Right Angles
Four multiple-choice items in this exercise all reduce to the same underlying fact — that every tangent meets its radius at exactly 90° — combined with one other angle rule each time.
| Given | Reasoning | Answer |
|---|---|---|
| Angle between a tangent and the radius at the point of contact | Definition, no calculation needed | 90° |
| Tangent length 24 cm, distance from centre to external point 25 cm | 25²=r²+24² ⟹ r²=49 | 7 cm |
| ∠POQ=110° between two radii to the tangent points | Quadrilateral OAPB's angles sum to 360°, two are 90° each ⟹ remaining two sum to 180° | ∠PAQ = 70° |
| ∠APB=80° between two tangents from P | ∠PAO=90°, OP bisects ∠APB so ∠APO=40°, angle sum in △APO | ∠POA = 50° |
The third row's 360°-minus-two-right-angles shortcut is worth remembering on its own: any time a problem gives the angle between two radii to a pair of tangent points, the angle between the two tangents themselves is just 180° minus that value — no need to redraw the whole quadrilateral each time to rederive it. The fourth row runs a related idea in the other direction, starting from the angle between the tangents instead of the angle between the radii, and reaching the same 180°-supplementary relationship through the angle-bisector fact noted above rather than the quadrilateral-angle-sum route.
A Chord Tangent to a Smaller Circle
Two concentric circles have radii 5 cm and 3 cm. A chord of the larger circle that just touches the smaller one is cut exactly in half at the point of contact, since a radius drawn perpendicular to a chord always bisects it.
OA² = OP² + AP² ⟹ 5² = 3² + AP² ⟹ AP = 4 ⟹ AB = 2×AP = 8 cmA Parallelogram Circumscribing a Circle Is Always a Rhombus
If all four sides of a parallelogram ABCD are tangent to a single inscribed circle, the Length of Tangents Theorem applied at all four corners forces something the parallelogram's own definition never guaranteed on its own: AP=AS, BP=BQ, CR=CQ, DR=DS. Adding all four equalities together and regrouping:
AB + CD = AD + BC ⟹ 2AB = 2AD (since AB=CD and AD=BC in any parallelogram) ⟹ AB = ADAdjacent sides equal in a parallelogram is exactly what defines a rhombus. Every parallelogram that circumscribes a circle is a rhombus — a fact that isn't true of parallelograms in general, only the ones built specifically to wrap around an inscribed circle. The rhombus's own two extra properties — equal sides all around, and diagonals crossing at right angles — never needed to be assumed anywhere in this proof; they simply fell out once AB=AD was established.
Solving a Triangle From Its Inscribed Circle
A triangle ABC circumscribes a circle of radius 3 cm, touching side BC at D, with BD=9 cm and DC=3 cm. Finding AB and AC starts the same way as the rhombus proof — equal tangent lengths from each vertex — but finishes with a clever shape spotted inside the figure.
Let AF = AE = x (equal tangents from A) ⟹ AB = x+9, AC = x+3, BC = 12Quadrilateral OECD (O the centre, E and D the two points of contact on AC and BC) has OE=OD (both radii) and CE=CD (both tangents from C), with right angles at E and D — enough to make OECD a square, forcing ∠C=90°. That right angle turns triangle ACB into a genuine Pythagoras problem:
(x+9)² = (x+3)² + 12² ⟹ 18x+81 = 6x+9+144 ⟹ 12x=72 ⟹ x=6
AB = 6+9 = 15 cm, AC = 6+3 = 9 cmA Tangent That Bisects a Side
In right triangle ABC, a circle drawn with AB as its diameter crosses hypotenuse AC at P. The claim: the tangent to that circle at P always bisects BC, wherever exactly P happens to land.
∠QPC = ∠QCP (given) ⟹ QP = QC
QP = QB (equal tangents from external point Q)
⟹ QC = QB — the tangent through P bisects BC at QThe two equal-length facts here come from entirely different sources — one from a given angle equality, the other from the Length of Tangents Theorem proved earlier in this exercise — and only meet because both happen to equal the same segment QP, the shared piece that ties the whole proof together.
Four Constructions, One Hands-On Skill
The remaining problems in this exercise step away from proof entirely: constructing a pair of tangents from a point 10 cm from a 6 cm-radius circle and verifying the length by Pythagoras; drawing a tangent from a point on one circle to a smaller concentric circle; tracing a circle from a bangle and building tangents from an external point to it; and simply counting how many tangents reach a circle from one outside point (always two, confirmed by construction rather than proof this time). Every one of these turns the theorem proved earlier in this exercise into something drawn and measured, not just argued algebraically.
From Tangent Pairs to Secant Segments
This exercise's every result came from a tangent touching a circle at a single point. Exercise 9.3 turns to the opposite case — a secant's chord, and the two curved regions of area it slices the circle's interior into. For the tangent-radius right angle every proof in this exercise depended on, revisit the chapter introduction.