Goals for today

Today we will discuss:

(a)

tangents to circles and tangent circles;

(b)

the relation between tangency and perpendicularity to the radius;

(c)

the alignment between the centers of tangent circles and the tangency point;

(d)

the angle at the center and the angle at the circumference;

(e)

angles in the same segment and the angle in a semicircle;

(f)

the tangent-chord theorem (Proposition III.32).

Tangents

Definition

We say that a line touches or is tangent to a circle if it intersects the circle at a single point.

We also say that two circles touch or are tangent to each other if they intersect in a single point.

There are two possibilities:

• the circles are tangent externally;
• the circles are tangent internally.

A basic fact about tangent circles

Proposition III.11–III.12. If two circles are tangent, then their centers and the point of tangency are aligned.

Proof. See Propositions III.11 and III.12.

Tangency and perpendicularity

Proposition. Consider a line \(\ell \) intersecting a circle of center \(C\) at a point \(A\). Then

Proof. We only prove the converse implication. For the direct implication, see Proposition III.16, and also Proposition III.18.

Assume that

We want to prove that \(A\) is the only intersection point of \(\ell \) with the circle.

Proceed by contradiction and assume that \(\ell \) intersects the circle at both \(A\) and \(B\).

Since

triangle \(CAB\) is isosceles. Hence the angles at \(A\) and \(B\) are equal. But \(\angle A\) is a right angle, since \(\ell \perp CA\). Therefore \(\angle B\) is also a right angle.

So

But then

which contradicts Proposition I.32.

This contradiction shows that \(A\) is the only point where \(\ell \) meets the circle. Therefore \(\ell \) is tangent to the circle at \(A\).

Application

Consider a circle \(\mathcal {C}\), and assume that \(AB\) and \(AC\) are tangent to \(\mathcal {C}\) at \(B\) and \(C\), respectively. Show that

Angles in circles

Angle at the center and angle at the circumference

Proposition III.20. Consider a circle of center \(C\), a chord \(AB\), and a point \(D\) on the circle on the same side of the line \((AB)\) as \(C\). Then

Proof. We need to show that

Draw the radius \(CD\), and split the central angle:

Since

triangle \(ACD\) is isosceles, so

By Proposition I.32 in triangle \(ACD\),

Since the first two angles are equal, this gives

Equivalently,

Similarly, since

triangle \(CDB\) is isosceles, so the same argument gives

Therefore

Application: equal angles in the same segment

Proposition III.21. Consider a chord \(AB\) of a circle of center \(C\), and points \(D\) and \(E\) on the circle on the same side of \(C\) with respect to the line \((AB)\). Show that

Application: angle in a semicircle

Proposition III.31. Let \(AB\) be a diameter of a circle and \(C\) a point on the circle. Show that

is a right angle.

The tangent-chord theorem

Proposition III.32. Let \(\ell \) be tangent to a circle \(\mathcal {C}\) at \(A\), and let \(AB\) be a chord. Then the angle between the tangent \(\ell \) and the chord \(AB\) is equal to the angle in the alternate segment.

In the notation of the figure:

Proof. Thanks to Proposition III.21, we may assume without loss of generality that \(AD\) is a diameter of the circle.

We prove first that

By Proposition I.32 applied to triangle \(ADB\),

But since \(AD\) is a diameter, Proposition III.31 gives

Hence

On the other hand, since \(\ell \) is tangent at \(A\), Proposition III.18 gives

Also,

Combining (1), (2), and (3), we get

Therefore

We leave it to the reader, using Proposition III.32 in the same spirit, to show that