The three kinds of line.
chord
joins two points on the circle
secant
a chord extended out
tangent
touches at one point only
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Chords, Tangents and Secants
Lines meeting a circle — and the products that always come out equal.
Find x
Two chords cross. One is split into 6 and 4; the other into 3 and x. Find x.
Where two chords cross, the two products match: 6 × 4 = 3 × x, so 24 = 3x and x = 8. B adds the pieces instead of multiplying them — the rule is about products, never sums.
Two facts about tangents.
| Fact | Use |
|---|---|
| tangent ⊥ radius | a right angle at the point of contact |
| two tangents from one point | are equal in length |
The perpendicular one hands you a right triangle — and with it, Pythagoras.
A tangent 12 units long from a point 13 from the centre gives a radius of √169 − 144 = 5, from the right angle alone.
Two tangents and the angle between them. Tangents PA and PB meet at P with ∠APB = 50°. The radii OA and OB each meet their tangent at 90°, so quadrilateral OAPB has angles 90°, 90°, 50° and ∠AOB, which total 360°: ∠AOB = 360 − 90 − 90 − 50 = 130°. The angle at the centre and the angle at P always add to 180°.
Two chords crossing inside.
a · b = c · d
the two pieces of one chord, times the two pieces of the other
6 × 4 = 3 × x
set the products equal
24 = 3x
tidy
x = 8
÷ 3
Pieces of the same chord multiply
6 and 4 belong to one chord; 3 and x to the other. Multiplying 6 by 3 instead mixes the chords and gives nonsense.
Lines meeting outside the circle.
Whole × outside, every time.
| Setup | Rule |
|---|---|
| two secants | whole₁ × outside₁ = whole₂ × outside₂ |
| tangent and secant | t² = whole × outside |
outside = 4, inside = 5
so the whole is 9
t² = 9 × 4
whole times outside
t² = 36, t = 6
the tangent length
“Whole” means the entire secant
Not just the part inside the circle. Using 5 instead of 9 gives t² = 20 — the single most common error in this topic.
One idea behind all three.
chords: a · b = c · d
from inside
secants: whole × outside, both ways
from outside
tangent: whole and outside are the same line
so t × t = t²
The tangent rule is not a separate formula — it is the secant rule where both distances happen to be equal.
Mathematicians call the shared value the power of the point. Every line drawn through that point produces the same product.
When x sits in both pieces: a quadratic.
A tangent from P has length 10. A secant from P has outside part x and inside part 15. Find x.
whole = x + 15, outside = x
x appears twice
x(x + 15) = 10² = 100
whole × outside = t²
x² + 15x − 100 = 0
set to zero
(x + 20)(x − 5) = 0
factors: 20 × 5 = 100, 20 − 5 = 15
x = 5
a length cannot be −20
Check: 5 × 20 = 100 = 10² ✓. The same move works for crossing chords with pieces x and x + k.
Do not multiply x by the inside part alone
15x = 100 treats the inside piece as the whole and gives 203. The whole secant is x + 15, so x must be multiplied by x + 15.
Variations: two secants, a tangent that is not a whole number.
Two secants from P meet the circle at A, B and at C, D. PA = 5, AB = 7, PC = 6 and CD = x. Find x.
wholes: PB = 5 + 7 = 12, PD = 6 + x
outside + inside
5 × 12 = 6(6 + x)
whole × outside, both secants
60 = 36 + 6x, x = 4
check: 6 × 10 = 60 ✓
Different: no tangent, so neither side is a square. Each side is its own secant’s outside part times its own whole.
Not outside × inside
5 × 7 = 6 × x gives x = 356, a fraction where a clean answer was waiting. The inside piece is never multiplied on its own from an outside point.
From P, a secant has outside part 3 and inside part 5, and a tangent has length t. Find t.
t² = (3 + 5) × 3 = 24
whole × outside
t = √24 = √4 × 6 = 2√6
≈ 4.9, between 3 and 8 ✓
Different: 24 is not a perfect square, so the answer stays as a simplified root.
Two favourite set-ups: a secant through the centre, and a circle inside a triangle.
A secant from P passes through the centre. Its outside part is 3, and the tangent from P is 9. Find the radius r. The inside part is a diameter, 2r.
9² = 3(3 + 2r)
t² = whole × outside
27 = 3 + 2r
÷ 3
r = 12
24 is the diameter — not the radius
A circle touches the sides of △ABC at D (on AB), E (on BC) and F (on CA). AB = 9, BC = 11, CA = 8. Find BD.
BD = BE = x
two tangents from B
AF = AD = 9 − x, CF = CE = 11 − x
the same rule from A and C
(9 − x) + (11 − x) = 8
AF + FC = CA
20 − 2x = 8, x = 6
BD = 6
On the exam
Expect the unknown in two places (a quadratic), a secant through the centre that hides the radius, two equal tangents written as expressions, or a circle inscribed in a triangle. Label every piece first, then decide: inside the circle, piece × piece; outside, whole × outside. Ten-second check: put your answer back and the two products must match.
Check: the products must match.
first chord
6 × 4 = 24
—
second chord
3 × 8 = 24
equal ✓
tangent
6² = 36
9 × 4 = 36 ✓
sensible?
t = 6 is between 4 and 9
it always is ✓
Now you try
Practice problem
Find y
Two chords cross at segments 5 and 9, and 6 and y. Find y.
Equal products again: 5 × 9 = 6 × y, so 45 = 6y and y = 7.5. C is the product left undivided. A non-whole answer is fine here — nothing says the segments must be integers.