Tangents and the Tan-Chord Theorem

Tangent theorems connect a line outside the circle to the angles inside it. The tan-chord theorem is one of the most used results in Paper 2 riders, and its proof is examinable.

Tangent perpendicular to radius

Theorem. A tangent to a circle is perpendicular to the radius at the point of contact. (Reason: tan $⊥$ radius)

Converse. A line perpendicular to a radius at its endpoint on the circle is a tangent. (Reason: converse tan $⊥$ radius)

Example. $P T$ is a tangent at $T$ to a circle with centre $O$ and radius 5 cm. If $O P = 13$ cm, then $\hat{O T P} = 90°$ and $P T = 1 3^{2} - 5^{2} = 12$ cm.

Tangents from a common point

Theorem. Two tangents drawn from the same point outside a circle are equal in length. (Reason: tans from common pt)

The proof uses congruent triangles: $△ O A P \equiv △ O B P$ (RHS), since $O A = O B$ (radii), $O P$ is common, and both triangles have a right angle at the point of contact.

Consequences:

$△ P A B$ is isosceles, so $\hat{P A B} = \hat{P B A}$ .
$O P$ bisects $\hat{A P B}$ and $\hat{A O B}$ .
Quadrilateral $P A O B$ is cyclic, because $\hat{A} + \hat{B} = 180°$ .

Example. If $\hat{A P B} = 50°$ , then $\hat{P A B} = \hat{P B A} = 65°$ and $\hat{A O B} = 130°$ .

The tan-chord theorem

Theorem. The angle between a tangent and a chord equals the angle in the alternate segment. (Reason: tan chord theorem)

Proof (examinable). Let $T A$ be the tangent at $A$ , $A B$ the chord and $C$ a point in the alternate segment. Draw diameter $A O D$ and join $D B$ .

$\hat{T A D} = 90°$ (tan $⊥$ radius), so $\hat{T A B} = 90° - \hat{B A D}$ .
$\hat{A B D} = 90°$ ( $∠$ in semi-circle), so $\hat{A D B} = 90° - \hat{B A D}$ (sum of $∠$ s in $△$ ).
So $\hat{T A B} = \hat{A D B}$ .
$\hat{A D B} = \hat{A C B}$ ( $∠$ s in the same seg).

Therefore $\hat{T A B} = \hat{A C B}$ .

Converse. If a line through the end of a chord makes an angle equal to the angle in the alternate segment, the line is a tangent.

Finding the alternate segment is the key skill. It is on the other side of the chord from the angle between the tangent and the chord.

Key terms

Tangent: A line that touches a circle at exactly one point.
Point of contact: The point where a tangent touches the circle.
Alternate segment: The segment on the opposite side of the chord from the tangent angle.
Tan chord theorem: The angle between a tangent and a chord equals the angle in the alternate segment.

Common mistakes

Using the angle in the wrong segment with the tan-chord theorem.
Assuming a line is a tangent without it being given or proved.
Forgetting that tangents from a common point make an isosceles triangle.
Leaving out the construction (diameter $A O D$ ) in the tan-chord proof.

Practice questions

Reveal steps one at a time, then tick the marks you earned.

Question 1

4 marks total

knowledge

$P T$ and $P S$ are tangents from $P$ to a circle with centre $O$ and radius 5 cm, touching the circle at $T$ and $S$ . $O P = 13$ cm.

Calculate the lengths of $P T$ and $P S$ , with reasons.

Worked solution

Hidden - try the question first

Solution

$\hat{O T P} = 90°$ (tan $⊥$ radius)
$P T^{2} = 1 3^{2} - 5^{2} = 144$ (Pythagoras)
$P T = 12$ cm
$P S = P T = 12$ cm (tans from common pt)

Question 2

5 marks total

knowledge

Prove the theorem that states that the angle between a tangent and a chord is equal to the angle in the alternate segment.

Worked solution

Hidden - try the question first

Solution

Construction: diameter $A O D$ ; join $D B$
$\hat{T A D} = 90°$ (tan $⊥$ radius) and $\hat{A B D} = 90°$ ( $∠$ in semi-circle)
$\hat{T A B} = 90° - \hat{B A D}$ and $\hat{A D B} = 90° - \hat{B A D}$ (sum of $∠$ s in $△$ )
So $\hat{T A B} = \hat{A D B}$
$\hat{A D B} = \hat{A C B}$ ( $∠$ s in the same seg), so $\hat{T A B} = \hat{A C B}$

Question 3

6 marks total

complex

$T A$ is a tangent to the circle at $A$ . $B$ and $C$ lie on the circle, with $C$ in the segment on the opposite side of chord $A B$ from $T$ . $\hat{T A B} = 58°$ and $\hat{A B C} = 64°$ .

(a) Calculate $\hat{A C B}$ , with a reason. (2)

(b) Calculate $\hat{B A C}$ . (2)

Worked solution

Hidden - try the question first

Solution

(a) $A O D$ (tan chord theorem)
$= 58°$
(b) $\hat{B A C} = 180° - 58° - 64°$ (sum of $∠$ s in $△$ )
$= 58°$
(c) $\hat{B A C} = \hat{A C B} = 58°$
$B A = B C$ (sides opp equal $∠$ s), so $△ A B C$ is isosceles

Question 4

9 marks total

problem-solving

$P A$ and $P B$ are tangents to a circle with centre $O$ at $A$ and $B$ . $C$ is a point on the major arc $A B$ . $\hat{A P B} = 50°$ .

(a) Calculate $\hat{P A B}$ . (2)

(b) Calculate $\hat{A C B}$ . (2)

(d) Calculate $\hat{A O B}$ . (2)

Worked solution

Hidden - try the question first

Solution

(a) $A O D$ (tans from common pt), so $\hat{P A B} = \hat{P B A}$
$\hat{P A B} = \frac{1}{2} (180° - 50°) = 65°$
(b) $\hat{A C B} = \hat{P A B}$ (tan chord theorem)
$= 65°$
(c) $\hat{O A P} = 90°$ and $\hat{O B P} = 90°$ (tan $⊥$ radius)
$\hat{O A P} + \hat{O B P} = 180°$
$P A O B$ is cyclic (converse opp $∠$ s of cyclic quad)
(d) $P T$ (opp $∠$ s of cyclic quad)
$\hat{A O B} = 130°$

FAQs

Is the tan-chord theorem proof examinable in IEB Grade 12?

Yes. It is one of the examinable theorem proofs, so learn the construction and each reason.

How do I find the alternate segment?

Look at the chord that makes the angle with the tangent. The alternate segment is on the other side of that chord.

Why is a quadrilateral with two tangents and two radii cyclic?

Both angles between a tangent and its radius are 90°, so those opposite angles add up to 180°.

Tangent perpendicular to radius

Tangents from a common point

The tan-chord theorem

Key terms

Common mistakes

Practice questions

FAQs

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