Tangents to a Circle

A tangent touches a circle at exactly one point. Two facts about tangents, that they are perpendicular to the radius and that two tangents from the same point are equal, appear in almost every circle geometry question from here on.

A tangent is perpendicular to the 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. A tangent is drawn from $P$ to a circle with radius 7 cm. If $P$ is 25 cm from the centre, the tangent length is $2 5^{2} - 7^{2} = 24$ cm.

Two tangents from a common point are equal

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

Proof. Tangents $P A$ and $P B$ touch the circle at $A$ and $B$ . Join $O A$ , $O B$ and $O P$ .

$\hat{A} = \hat{B} = 90°$ (tan $⊥$ radius)
$O A = O B$ (radii) and $O P$ is common
$△ O A P \equiv △ O B P$ (RHS), so $P A = P B$

The congruence also shows that $O P$ bisects $\hat{A P B}$ and $\hat{A O B}$ .

Inscribed circle. If a circle touches all three sides of a triangle, the tangent lengths from each vertex are equal. With sides 9, 10 and 11, the tangent lengths are 5, 4 and 6.

Key terms

Tangent: A line that touches a circle at exactly one point.
Point of contact: The point where a tangent touches the circle.
Inscribed circle: A circle inside a polygon that touches every side.

Common mistakes

Assuming a line is a tangent because it looks like one.
Using the tangent as the hypotenuse in Pythagoras; the hypotenuse is the line to the centre.
Forgetting that the two tangents from a point make an isosceles triangle with the chord joining the points of contact.

Practice questions

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

Question 1

3 marks total

knowledge

$P T$ is a tangent from $P$ to a circle with centre $O$ and radius 7 cm, touching it at $T$ . $O P = 25$ cm. Calculate $P T$ .

Worked solution

Hidden - try the question first

Solution

$\hat{O T P} = 90°$ (tan $⊥$ radius)
$P T^{2} = 2 5^{2} - 7^{2} = 576$
$P T = 24$ cm

Question 2

5 marks total

routine

$P A$ and $P B$ are tangents to a circle with centre $O$ at $A$ and $B$ . $\hat{A P B} = 70°$ . Calculate, with reasons:

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

(b) $\hat{O A B}$ (2)

Worked solution

Hidden - try the question first

Solution

(a) $P A = P B$ (tans from common pt), so $\hat{P A B} = \hat{P B A}$
$\hat{P A B} = \frac{1}{2} (180° - 70°) = 55°$
(b) $\hat{O A P} = 90°$ (tan $⊥$ radius)
$\hat{O A B} = 90° - 55° = 35°$
(c) $\hat{A O B} = 180° - 70° = 110°$ (sum of $∠$ s in quad $O A P B$ )

Question 3

6 marks total

complex

A circle is inscribed in $△ A B C$ and touches $A B$ at $P$ , $B C$ at $Q$ and $C A$ at $R$ . $A B = 9$ cm, $B C = 10$ cm and $C A = 11$ cm.

Calculate $A P$ , $B Q$ and $C R$ .

Worked solution

Hidden - try the question first

Solution

$\hat{O T P} = 90°$ , $B P = B Q = y$ , $C Q = C R = z$ (tans from common pt)
$x + y = 9$ , $y + z = 10$ , $x + z = 11$
Adding: $2 (x + y + z) = 30$ , so $x + y + z = 15$
$x = 15 - 10 = 5$ and $y = 15 - 11 = 4$
$z = 15 - 9 = 6$

Question 4

8 marks total

problem-solving

In the diagram, $P A$ and $P B$ are tangents to a circle with centre $O$ at $A$ and $B$ . $O P$ cuts $A B$ at $M$ .

(a) Prove that $O P$ bisects $\hat{A P B}$ . (4)

(b) Prove that $A B ⊥ O P$ . (4)

Worked solution

Hidden - try the question first

Solution

(a) $P A = P B$ (tan $⊥$ radius)
$O A = O B$ (radii); $O P$ common
$△ O A P \equiv △ O B P$ (RHS)
$\hat{A P O} = \hat{B P O}$ ( $\equiv$ triangles), so $⊥$ bisects $\hat{A P B}$
(b) In $△ A P M$ and $△ B P M$ : $P A = P B$ (tans from common pt)
$\hat{A P M} = \hat{B P M}$ (from (a)); $P M$ common
$△ A P M \equiv △ B P M$ (SAS), so $\hat{M}_{1} = \hat{M}_{2}$
$\hat{M}_{1} + \hat{M}_{2} = 180°$ , so each is $90°$ and $A B ⊥ O P$

FAQs

How do I prove two tangents from a point are equal?

Join the centre to the point and to both points of contact, then use RHS congruence.

What is the reason for a tangent meeting a radius at 90°?

Write 'tan $⊥$ radius'.

Where do tangent theorems appear in Grade 12?

In Euclidean geometry riders, the tan-chord theorem, and in analytical geometry when finding tangent equations.

A tangent is perpendicular to the radius

Two tangents from a common point are equal

Key terms

Common mistakes

Practice questions

FAQs

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