Geometry (B): READ: Chords Reading

Chords

Learning Objectives

Find the lengths of chords in a circle.
Find the measure of arcs in a circle.

Introduction

Chords are line segments whose endpoints are both on a circle. The figure shows an arc $\widehat{AB}$ and its related chord $\overline{AB}$ .

There are several theorems that relate to chords of a circle that we will discuss in the following sections.

Perpendicular Bisector of a Chord

Theorem The perpendicular bisector of a chord is a diameter.

Proof

We will draw two chords, $\overline{AB}$ and $\overline{CD}$ such that $\overline{AB}$ is the perpendicular bisector of $\overline{CD}$ .

We can see that $\triangle COE \cong \triangle DOE$ for any point $O$ on chord $\overline{AB}$ .

The congruence of the triangles can be proven by the $SAS$ postulate:

$\overline{CE} & \cong \overline{ED} \\ \overline{OE} & \cong \overline{OE}$

$\angle{OEC}$ and $\angle{OED}$ are right angles

This means that $\overline{CO} \cong \overline{DO}$ .

Any point that is equidistant from $C$ and $D$ lies along $\overline{AB}$ , by the perpendicular bisector theorem. Since the center of the circle is one such point, it must lie along $\overline{AB}$ so $\overline{AB}$ is a diameter.

If $O$ is the midpoint of $\overline{AB}$ then $\overline{OC}$ and $\overline{OD}$ are radii of the circle and $\overline{AB}$ is a diameter of the circle. $\blacklozenge$

Perpendicular Bisector of a Chord Bisects Intercepted Arc

Theorem The perpendicular bisector of a chord bisects the arc intercepted by the chord.

Proof

We can see that $\triangle CAD \cong \triangle BAD$ because of the $SAS$ postulate.

$\overline{DA} & \cong \overline{DA} \\ \overline{BD} & \cong \overline{DC}$

$\angle{ADB}$ and $\angle{ADC}$ are right angles.

This means that $\angle{DAB} \cong \angle{DAC} \Rightarrow \widehat{BE} \cong \widehat{CE}$ .

This completes the proof. $\blacklozenge$

Congruent Chords Equidistant from Center

Theorem Congruent chords in the same circle are equidistant from the center of the circle.

First, recall that the definition of distance from a point to a line is the length of the perpendicular segment drawn to the line from the point. $CO$ and $DO$ are these distances, and we must prove that they are equal.

Proof.

$\triangle AOE \cong \triangle BOF$ by the SSS Postulate.

$\overline{AE} & \cong \overline{BF}\ \text{(given)} \\ \overline{AO} & \cong \overline{BO}\ \text{(radii)}\\ \overline{EO} & \cong \overline{OF}\ \text{(radii)}$

Since the triangles are congruent, their corresponding altitudes $\overline{OC}$ and $\overline{OD}$ must also be congruent: $\overline{CO} \cong \overline{DO}$ .

Therefore, $\overline{AE} \cong \overline{BF}$ and are equidistant from the center. $\blacklozenge$

Converse of Congruent Chords Theorem

Theorem Two chords equidistant from the center of a circle are congruent.

This proof is left as a homework exercise.

Next, we will solve a few examples that apply the theorems we discussed.

Example 1

$CE = 12\;\mathrm{inches}$ , and is $3\;\mathrm{in}$ . from the center of circle $O$ .

A. Find the radius of the circle.

B. Find $m\widehat{CE}$

Draw the radius $\overline{OC}$ .

A. $\overline{OC}$ is the hypotenuse of the right triangle $\triangle COT$ .

$OT = 3\;\mathrm{in}$ .; $CT = 6\;\mathrm{in}$ .

Apply the Pythagorean Theorem.

$(OC)^2 &= (OT)^2 + (CT)^2\\ (OC)^2 &= 3^2 + 6^2 = 45\\ (OC)^2 &= 3 \sqrt{5} \approx 6.7\ \text{in.}$

B. Extend $\overline{OT}$ to intersect the circle at point $D$ .

$m\widehat{CE} &= 2 m\widehat{CD}\\ m\widehat{CD} &= m\angle{COD}\\ tan \angle{COD} &= \frac{6} {3} = 2\\ m\angle{COD} &\approx 63.4^\circ\\ m\widehat{CE} &\approx 126.9^\circ$

Example 2

Two concentric circles have radii of $6\;\mathrm{inches}$ and $10\;\mathrm{inches.}$ A segment tangent to the smaller circle is a chord of the larger circle. What is the length of the segment?

Start by drawing a figure that represents the problem.

$OC & = 6\ \text{in} \\ OB & = 10\ \text{in}.$

$\triangle COB$ is a right triangle because the radius $\overline{OC}$ of the smaller circle is perpendicular to the tangent $\overline{AB}$ at point $C$ .

Apply the Pythagorean Theorem.

$(OB)^2 &= (OC)^2 + (BC)^2\\ 10^2 &= 6^2 + BC^2\\ BC &= 8\ \text{in}\\ AB &= 2BC\ \text{from Theorem 9.6}\\ \text{Therefore,} AB &= 16 \text{in.}$

Example 3

Find the length of the chord of the circle.

$x^2 + y^2 = 9$ that is given by line $y = -2x- 4$ .

First draw a graph that represents the problem.

Find the intersection point of the circle and the line by substituting for $y$ in the circle equation.

$x^2 + y^2 &= 9\\ y &= -2x -4\\ x^2 + ( -2x - 4)^2 &= 9\\ x^2 + 4 x^2 + 16x + 16 &= 9\\ 5x^2 + 16x + 7 &= 0$

Solve using the quadratic formula.

$x = -0.52$ or $-2.68$

The corresponding values of $y$ are

$y = - 2.96$ or $1.36$

Thus, the intersection points are approximately $(-0.52, -2.96)$ and $(-2.68, 1.36)$ .

We can find the length of the chord by applying the distance formula:

$d = \sqrt{(-0.52+2.68)^2 + (-2.96 - 1.36)^2} \approx 4.83\ \text{units}.$

Example 4

Let $A$ and $B$ be the positive $x-$ intercept and the positive $y-$ intercept, respectively, of the circle $x^2 + y^2 = 32$ . Let $P$ and $Q$ be the positive $x-$ intercept and the positive $y-$ intercept, respectively, of the circle $x^2 + y^2 = 64$ .

Verify that the ratio of chords $AB : PQ$ is the same as the ratio of the corresponding diameters.

For the circle $x^2 + y^2 = 32$ , the $x-$ intercept is found by setting $y = 0$ . So $A = (4 \sqrt{2},0)$ .

The $y-$ intercept is found by setting $x = 0$ . So, $B = (0, 4 \sqrt{2})$ .

$AB$ can be found using the distance formula: $AB = \sqrt{(4 \sqrt {2}^2) + (4 \sqrt{2})^2} = \sqrt{32 + 32} = 8.$

For the circle $x^2 + y^2 = 64$ , $P = (8 , 0)$ and $Q = (0 , 8)$ .

$PQ = \sqrt{(8)^2 + (8)^2} = \sqrt{64 + 64} = 8\sqrt{2}.$

The ratio of the $AB:PQ = 8: 8\sqrt{2} = 1:\sqrt{2}$ .

Diameter of circle $x^2 + y^2 = 32$ is $8 \sqrt{2}$ .

Diameter of circle $x^2 + y^2 = 64$ is $16$ .

The ratio of the diameters is $8 \sqrt{2}: 16 = 1: \sqrt{2}$

The ratio of the chords and the ratio of the diameters are the same.

Lesson Summary

In this section we gained more tools to find the length of chords and the measure of arcs. We learned that the perpendicular bisector of a chord is a diameter and that the perpendicular bisector of a chord also bisects the corresponding arc. We found that congruent chords are equidistant from the center, and chords equidistant from the center are congruent.

The questions are for your own review. The answers are listed below to help you check your work and understanding.

Review Questions

Find the value of :
Find the measure of .
Two concentric circles have radii of $3\;\mathrm{inches}$ and $8\;\mathrm{inches.}$ A segment tangent to the smaller circle is a chord of the larger circle. What is the length of the segment?
Two congruent circles intersect at points $A$ and $B$ . $\overline{AB}$ is a chord to both circles. If the segment connecting the centers of the two circles measures $12\;\mathrm{in}$ and $\overline{AB}= 8\;\mathrm{in}$ , how long is the radius?
Find the length of the chord of the circle $x^2 + y^2 = 16$ that is given by line $y = x + 1$ .
Prove Theorem 9-9.
Sketch the circle whose equation is $x^2 + y^2 = 16$ . Using the same system of coordinate axes, graph the line $x + 2y = 4$ , which should intersect the circle twice—at $A = (4, 0)$ and at another point $B$ in the second quadrant. Find the coordinates of $B$ .
Also find the coordinates for a point $C$ on the circle above, such that $\overline{AB} \cong \overline{AC}$ .
The line $y = x + 1$ intersects the circle $x^2 + y^2 = 9$ in two points. Call the third quadrant point $A$ and the first quadrant point $B$ , and find their coordinates. Let $D$ be the point where the line through $A$ and the center of the circle intersects the circle again. Show that $\triangle{BAD}$ is a right triangle.
A circular playing field $100\;\mathrm{meters}$ in diameter has a straight path cutting across it. It is $25\;\mathrm{meters}$ from the center of the field to the closest point on this path. How long is the path?

Review Answers

1. $12.53$
2. $6.70$
3. $14.83$
4. $11.18$
5. $16$
6. $11.18$
7. $16.48$
8. $32$
1. $136.4^\circ$
2. $120^\circ$
3. $60^\circ$
4. $118.07^\circ$
5. $115^\circ$
6. $73.74^\circ$
7. $146.8^\circ$
8. $142.5^\circ$
$14.83$
$7.21$
$7.88$
proof
$(-12/5,16/5)$
$(-12/5,-16/5)$
$B(1.56,2.56)$ ; $A (-2.56,-1.56)$ ; $D (2.56, 1.56)$
$(AB)^2= 34$ ; $(AD)^2 = 36$ ; $(BD)^2 = 2$ ; $(AD)^2 = (AB)^2 + (BD)^2$
$86.6\;\mathrm{meters}$

Last modified: Tuesday, June 29, 2010, 11:36 AM