Chapter 9.2 — Exercise 9.1 — Length of Tangent
Length of the tangent drawn from an external point to a circle. This is Lesson 2 of 4 in Chapter 9: Tangents and Secants to a Circle.
Vocabulary First, Then One Formula
Exercise 9.1 opens by pinning down the exact words this whole chapter depends on, then spends the rest of its problems putting the tangent-length formula from the introduction to work — plus one proof that reaches beyond a single tangent to a pair of them. Getting the vocabulary genuinely precise matters more than it might seem, since several later problems hinge on knowing exactly which word describes which configuration without having to stop and puzzle it out.
Five Blanks Worth Getting Precise
| Statement | Answer |
|---|---|
| A tangent to a circle intersects it in ___ point(s). | One |
| A line intersecting a circle in two points is called a ___. | Secant |
| A circle can have ___ parallel tangents at the most. | Two |
| The common point of a tangent and the circle is called the ___. | Point of contact |
| We can draw ___ tangents to a circle. | Infinite |
The last two rows look like they could contradict each other, but they don't — they're answering genuinely different questions. A circle as a whole has infinitely many tangent lines, one at every single point around its edge. A single external point, by contrast, can only ever reach two of those tangent lines (a fact Exercise 9.2 proves directly). "Infinite" and "two" are both correct simultaneously, because one counts tangents to the whole circle and the other counts tangents from one specific point.
The third row is worth a second look too: "at the most" is doing real work in that sentence. Two parallel tangents exist only for one very specific pair of points on the circle — the two ends of a diameter, exactly the configuration proved later in this same exercise. Any other pair of tangent lines drawn on a circle will eventually meet if extended far enough, since two lines are parallel only in that one special case, not by default, however the two points happen to be chosen.
The Formula, Applied Twice
Both numeric problems in this exercise use the identical tangent-length formula from the chapter introduction, just with the radius and centre-distance swapped between which one is "given" and which is "found." Neither problem gives away in words which measurement is the radius and which is the centre-distance — that has to be inferred correctly from the description before the formula can even be set up, since the two numbers alone don't announce their own roles.
OP² = OA² + PA² ⟹ PA = √(OP² − OA²)| Given | Working | Tangent length |
|---|---|---|
| Radius = 5 cm, distance from centre to external point = 13 cm | PQ² = 13²−5² = 169−25 = 144 | 12 cm |
| Radius = 9 cm, distance from centre to external point = 15 cm | PQ² = 15²−9² = 225−81 = 144 | 12 cm |
Both problems land on the identical answer, 12 cm, despite starting from completely different radii and distances — a coincidence worth noticing rather than a sign something's wrong. 5-12-13 and 9-12-15 are both genuine Pythagorean triples (the second is just 3-4-5 scaled up by 3), and it happens that both share the same middle value. Confirming a result is a recognisable Pythagorean triple, the way both of these are, is a fast way to sanity-check a tangent-length answer before moving on.
It's worth being deliberate about which length plays which role in the formula, since mixing them up is the single most common error here. OP is always the distance from the external point to the centre — the longest of the three lengths, since it's the hypotenuse of the right triangle. OA is always the radius. PA, the one being solved for, is always the shorter leg. Sketching the right triangle quickly, even roughly, before substituting numbers is often enough to catch a radius and a centre-distance swapped by mistake, since the hypotenuse should visibly be the longest side once it's drawn — a check that takes seconds but catches a genuinely common substitution error.
Proving Two Tangents Parallel
Let AB be a diameter of a circle, with a tangent drawn at each end — PQ at A, RS at B. The claim: these two tangents are always parallel to each other, no matter how the diameter is oriented or how the circle itself is sized.
∠OAQ = 90° and ∠OBS = 90° (tangent ⊥ radius at each point of contact)
⟹ ∠OAQ + ∠OBS = 180°AB is a single straight line (the diameter) crossing both tangent lines, making it a transversal. Since the two angles it forms with the tangents on the same side add up to exactly 180° — co-interior angles summing to a straight angle — that's precisely the condition that forces two lines to be parallel. PQ ∥ RS. Every part of this proof traces back to the same single fact the whole chapter opened with: the tangent-radius right angle, guaranteed at both points of contact independently, and here doing double duty at once.
It's worth noticing what this proof does not need at all: no lengths, no circle radius, no coordinates. The entire argument runs purely on angle facts — a guaranteed right angle at each end of the diameter, and the co-interior-angles test for parallel lines — which is exactly why the conclusion holds for every circle and every diameter, not just one drawn to particular measurements. A proof that never touches a specific number is often the strongest kind, since there's nothing left to check by plugging in different values; it's simply true in general, for every circle that could ever be drawn.
A Construction, Left to the Compass
One further problem in this exercise breaks from proof and calculation entirely: draw a circle, then draw two lines parallel to a given line outside it, positioned so that one becomes a tangent and the other a secant. Like the constructions found throughout the previous chapter, this tests an entirely hands-on skill — placing a line at the exact right distance from the centre to either graze the circle once or cut through it twice — rather than anything solved with algebra on paper.
The construction is really a physical demonstration of the distance-versus-radius comparison introduced in the chapter's opening pages: a line parallel to the given one, moved gradually closer to the circle's centre, passes through non-intersecting, then tangent, then secant, exactly in that order as its distance from the centre shrinks past the radius. Drawing both a tangent and a secant version side by side, both parallel to the same original line, makes that transition visible on paper rather than just stated as a rule.
From One Tangent to a Matching Pair
Every problem here dealt with a single tangent line at a time. Exercise 9.2 moves to what happens once two tangents are drawn from the very same external point — proving they're always equal in length, working out the angles the resulting figure locks into place, and constructing the pair directly. That shift, from a single tangent's length to a pair of tangents' shared properties, is a genuinely new kind of question, not just a repeat of this exercise's Pythagoras substitution with a second triangle bolted on. For the tangent-length formula this exercise leaned on twice, revisit the chapter introduction.