The tangent to a circle

Geometry, grade 10 (5 units) Lesson 22 of 26Learning

The tangent to a circle

The tangent and the radius

A tangent to a circle is a line with exactly one point in common with the circle, the point of tangency. A line that cuts the circle at two points is a secant.

Theorem: A tangent to a circle is perpendicular to the radius at the point of tangency.

The proof is by contradiction. Suppose the tangent at $\displaystyle A$ is not perpendicular to $\displaystyle OA$ :

then the perpendicular from $\displaystyle O$ to the tangent meets it at another point, $\displaystyle M$ , and $\displaystyle OM<OA$ – a leg is shorter than the hypotenuse;
so $\displaystyle M$ is inside the circle, and the point $\displaystyle A'$ on the tangent with $\displaystyle {MA'}={MA}$ beyond $\displaystyle M$ is at the distance $\displaystyle OA$ from $\displaystyle O$ too: a second common point – which a tangent does not have.

The converse theorem: a line perpendicular to a radius at its end on the circle is a tangent: every other point of it is the end of a hypotenuse from $\displaystyle O$ , farther than the radius.

Two tangents from one point

Theorem: Two tangents to a circle from the same point are equal – the segments from the point to the points of tangency.

Given: $\displaystyle PA$ and $\displaystyle PB$ are tangents to the circle with centre $\displaystyle O$ , at $\displaystyle A$ and $\displaystyle B$ .
To prove: $\displaystyle {PA}={PB}$ .

The proof below draws $\displaystyle OP$ and shows that the right-angled triangles $\displaystyle OAP$ and $\displaystyle OBP$ are congruent:

	Claim	Reason
1	$\displaystyle {OA}={OB}$	Radii of the circle
2	$\displaystyle {\angle OAP}={\angle OBP}={90^{\circ}}$	A tangent is perpendicular to the radius at the point of tangency
3	$\displaystyle {OP}={OP}$	A common side
4	$\displaystyle \triangle OAP\cong\triangle OBP$	(1), (2), (3); two sides and the angle opposite the longer of them
5	$\displaystyle {PA}={PB}$	(4); corresponding sides of congruent triangles

From the same congruence: the segment from the centre to the point the tangents come from bisects the angle between the tangents (Exercise 2).

Exercise 1

$\displaystyle PA$ is a tangent to the circle with centre $\displaystyle O$ , at $\displaystyle A$ .
The radius is $\displaystyle 5$ , and $\displaystyle {OP}={13}$ .

Find

\displaystyle PA

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

\displaystyle PB

is a second tangent from

\displaystyle P

, and

\displaystyle {\angle APB}={50^{\circ}}

. Find

\displaystyle \angle AOB

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 2

$\displaystyle PA$ and $\displaystyle PB$ are tangents to the circle with centre $\displaystyle O$ , at $\displaystyle A$ and $\displaystyle B$ .
Prove: $\displaystyle {\angle APO}={\angle BPO}$ .

The angle between a tangent and a chord

Theorem: The angle between a tangent and a chord from the point of tangency equals the inscribed angle standing on the chord from its other side.

\displaystyle \angle BAT

\displaystyle =

\displaystyle \angle ACB

Given: $\displaystyle AT$ is a tangent to the circle with centre $\displaystyle O$ at $\displaystyle A$ ; $\displaystyle AB$ is a chord; $\displaystyle C$ is on the circle, on the other side of $\displaystyle AB$ .
To prove: $\displaystyle {\angle BAT}={\angle ACB}$ .

The proof below draws the diameter $\displaystyle AD$ . Both angles turn out to complete the angle $\displaystyle DAB$ to $\displaystyle 90^{\circ}$ :

	Claim	Reason
1	$\displaystyle {\angle DAT}={90^{\circ}}$	A tangent is perpendicular to the radius at the point of tangency
2	$\displaystyle {\angle ABD}={90^{\circ}}$	An inscribed angle on the diameter $\displaystyle AD$
3	$\displaystyle {\angle BAT}={90^{\circ}-\angle DAB}$	(1)
4	$\displaystyle {\angle ADB}={90^{\circ}-\angle DAB}$	(2); the sum of the angles of the triangle $\displaystyle ABD$
5	$\displaystyle {\angle ADB}={\angle ACB}$	Inscribed angles on the same arc $\displaystyle AB$
6	$\displaystyle {\angle BAT}={\angle ACB}$	(3), (4), (5)

Exercise 3

$\displaystyle AT$ is a tangent to the circle with centre $\displaystyle O$ at $\displaystyle A$ , and $\displaystyle AB$ is a chord.
$\displaystyle {\angle BAT}={40^{\circ}}$ , and $\displaystyle C$ is on the circle, on the other side of $\displaystyle AB$ .

Find

\displaystyle \angle ACB

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Find the central angle

\displaystyle \angle AOB

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 4

The lines $\displaystyle AP$ and $\displaystyle BQ$ are tangents to the circle with centre $\displaystyle O$ , at $\displaystyle A$ and at $\displaystyle B$ .
$\displaystyle AP\parallel BQ$ .
Prove: $\displaystyle {\angle AOB}={180^{\circ}}$ – the segment $\displaystyle AB$ is a diameter.

Exercise 5

Describe how to construct with a ruler and compasses, and explain why each construction works:

the tangent to a circle at a point $\displaystyle A$ on it;
the two tangents to a circle from a point $\displaystyle P$ outside it. (A hint: the circle with diameter $\displaystyle OP$ , and the inscribed angle on a diameter.)

Formulas: geometry, grade 10 (5 units)

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Introduction to analytic geometry

The distance between the points $\displaystyle (x_1,y_1)$ and $\displaystyle (x_2,y_2)$
$\displaystyle d$ $\displaystyle =$ $\displaystyle \sqrt{(x_1-x_2)^{2}+(y_1-y_2)^{2}}$
The midpoint of the segment
$\displaystyle M\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)$
The slope of a line through two points
$\displaystyle m$ $\displaystyle =$ $\displaystyle \frac{y_2-y_1}{x_2-x_1}$ $\displaystyle x_1\neq x_2$
The line of slope $\displaystyle m$ through the point $\displaystyle (x_1,y_1)$
$\displaystyle y-y_1$ $\displaystyle =$ $\displaystyle m(x-x_1)$
The explicit equation of a line – $\displaystyle m$ the slope, $\displaystyle b$ where it crosses the $\displaystyle y$ -axis
$\displaystyle y=mx+b$
The general equation of a line
$\displaystyle Ax+By+C=0$
Parallel lines
$\displaystyle m_1$ $\displaystyle =$ $\displaystyle m_2$
Perpendicular lines
$\displaystyle m_1\cdot m_2$ $\displaystyle =$ $\displaystyle -1$
A canonical circle – its centre at the origin, $\displaystyle R$ its radius
$\displaystyle x^{2}+y^{2}=R^{2}$
The circle with centre $\displaystyle (a,b)$ and radius $\displaystyle R$
$\displaystyle (x-a)^{2}+(y-b)^{2}=R^{2}$

Midsegments and the meeting point of the medians

A triangle’s midsegment – $\displaystyle DE$ is parallel to $\displaystyle BC$
$\displaystyle DE$ $\displaystyle =$ $\displaystyle \frac{1}{2}BC$
A trapezoid’s midsegment – $\displaystyle a$ , $\displaystyle b$ the bases
$\displaystyle EF$ $\displaystyle =$ $\displaystyle \frac{a+b}{2}$
The meeting point of the medians $\displaystyle G$ divides the median $\displaystyle AD$ in the ratio $\displaystyle 2:1$
$\displaystyle AG$ $\displaystyle =$ $\displaystyle 2GD$
Pythagoras’ theorem and its converse: the triangle is right-angled, the right angle opposite $\displaystyle c$
$\displaystyle a^{2}+b^{2}=c^{2}$

Proportion and similarity

A property of a proportion
$\displaystyle \frac{a}{b}=\frac{c}{d}\ \Rightarrow\ \frac{a+b}{b}=\frac{c+d}{d}$
Thales’ theorem – $\displaystyle DE\parallel BC$
$\displaystyle \frac{AD}{DB}$ $\displaystyle =$ $\displaystyle \frac{AE}{EC}$
The extended Thales theorem – $\displaystyle DE\parallel BC$
$\displaystyle \frac{AD}{AB}=\frac{AE}{AC}=\frac{DE}{BC}$
The angle bisector theorem – $\displaystyle AD$ bisects the angle $\displaystyle A$
$\displaystyle \frac{BD}{DC}$ $\displaystyle =$ $\displaystyle \frac{AB}{AC}$
The similarity ratio – corresponding sides
$\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=k$
Heights, medians, perimeters, the radii of the circumscribed and of the inscribed circles
$\displaystyle \frac{h_1}{h_2}=\frac{m_1}{m_2}=\frac{P_1}{P_2}=\frac{R_1}{R_2}=\frac{r_1}{r_2}=k$
The ratio of the areas
$\displaystyle \frac{S_1}{S_2}$ $\displaystyle =$ $\displaystyle k^{2}$

The circle

An inscribed angle is half the central angle on the same arc
$\displaystyle \angle ACB$ $\displaystyle =$ $\displaystyle \frac{1}{2}\angle AOB$
An inscribed angle on the diameter $\displaystyle AB$
$\displaystyle \angle ACB=90^{\circ}$
A quadrilateral $\displaystyle ABCD$ inscribed in a circle – opposite angles
$\displaystyle \angle A+\angle C$ $\displaystyle =$ $\displaystyle 180^{\circ}$
The angle between the tangent $\displaystyle AT$ and the chord $\displaystyle AB$ equals an inscribed angle on $\displaystyle AB$ from its other side
$\displaystyle \angle BAT$ $\displaystyle =$ $\displaystyle \angle ACB$
Two tangents from the point $\displaystyle P$
$\displaystyle PA=PB$
A quadrilateral $\displaystyle ABCD$ circumscribing a circle
$\displaystyle AB+CD$ $\displaystyle =$ $\displaystyle BC+AD$
The chords $\displaystyle AB$ and $\displaystyle CD$ meet at the point $\displaystyle E$
$\displaystyle AE\cdot EB$ $\displaystyle =$ $\displaystyle CE\cdot ED$
Two secants from the point $\displaystyle P$
$\displaystyle PA\cdot PB$ $\displaystyle =$ $\displaystyle PC\cdot PD$
A tangent $\displaystyle PT$ and a secant $\displaystyle PAB$ from the point $\displaystyle P$
$\displaystyle PT^{2}$ $\displaystyle =$ $\displaystyle PA\cdot PB$