Ninth Grade Mathematics: Equal Chords and Distance from Center
The length of a chord in a circle is inversely related to its distance from the center: as a chord moves closer to the center, it lengthens, reaching its maximum size at the diameter. Conversely, chords that are situated at the exact same distance from the center must have identical lengths. In this chapter, you will master the theorems governing equidistant chords and their geometric applications.
Equal Chords are Equidistant from the Center
The primary theorem connects the equality of chord lengths to their radial distance from the circle's center:
The Equidistant Chords Theorem:
Equal chords of a circle (or of congruent circles) are equidistant
from the center.
In Circle C(O, r):
If Chord AB = Chord CD,
then Perpendicular Distance OL = Perpendicular Distance OM.
. - ~ ~ - .
. ' O ' .
/ /|\ \
/ r / | \ r \
A ; / |OL\ ; C
*----|--------*---+---*-------|--*
| \ L | M / |
| \ | / |
B . _ | _ ' D
' - ~ ~ - .
If AB == CD ===> OL == OM
Deductive Proof of the Theorem
We prove this theorem using the RHS Congruence Criterion:
Given: A circle with center O and two chords AB and CD such that AB = CD.
OL is perpendicular to AB, and OM is perpendicular to CD.
To Prove: OL = OM.
Proof:
Step 1: The perpendicular from the center to a chord bisects the chord:
AL = (1/2) * AB and CM = (1/2) * CD.
Step 2: Since AB = CD (Given):
(1/2) * AB = (1/2) * CD
AL = CM.
Step 3: Connect the center to endpoints A and C to form radii OA and OC.
Consider right-angled Triangle OLA and Triangle OMC:
- Angle OLA = Angle OMC = 90 degrees (Perpendiculars)
- Hypotenuse OA = Hypotenuse OC = r (Radii of the same circle)
- Leg AL = Leg CM (Proven in Step 2)
Step 4: By the RHS Congruence Criterion:
Triangle OLA =~ Triangle OMC.
Step 5: By CPCTC:
OL = OM.
The proof is complete: equal chords are always equidistant from the center.
The Converse Theorem
The converse statement is equally true and frequently applied:
Converse Theorem:
Chords equidistant from the center of a circle are equal in length.
In Circle C(O, r):
If Perpendicular Distance OL = Perpendicular Distance OM,
then Chord AB = Chord CD.
Proof: In right triangles OLA and OMC, hypotenuse OA = OC = r, and leg OL = OM (given). By RHS congruence, Triangle OLA =~ Triangle OMC. By CPCTC, leg AL = CM. Multiplying both sides by 2 yields 2 * AL = 2 * CM. Since the perpendicular bisects the chord, 2AL = AB and 2CM = CD, proving AB = CD.
Intersecting Equal Chords Theorem
When two equal chords intersect inside a circle, their intersection point divides them into corresponding equal segments:
Theorem:
If two equal chords of a circle intersect within the circle, the segments
of one chord are equal to the corresponding segments of the other chord.
If Chord AB = Chord CD intersecting at point X:
AX = CX and XB = XD
A . - ~ ~ - .
* ' ' * C
/ \ / \
/ \ X / \
; *---*---* ;
| \ /O\ / |
\ * * /
\ / \ /
B . _ * * _ ' D
Proof: Drop perpendiculars OL to AB and OM to CD. Join OX. Since AB = CD, OL = OM (equal chords are equidistant). In right triangles OLX and OMX: Angle OLX = Angle OMX = 90 deg, hypotenuse OX = OX (common), and leg OL = OM. By RHS congruence, Triangle OLX =~ Triangle OMX. By CPCTC, LX = MX. Since AL = CM (half of equal chords), adding equals gives AL + LX = CM + MX, which yields AX = CX. Subtracting this from AB = CD yields XB = XD.
Chapter Exercises
Exercise 1: In a circle of radius 5 cm, two chords AB and CD are each of length 8 cm. Calculate their distances from the center, and state whether they are equidistant.
Exercise 2: Two equal chords AB and CD of a circle with center O intersect at point X inside the circle. Prove that the line segment joining the center O to the intersection point X bisects the angle between the two chords.
Exercise 3: In a circle of radius 13 cm, chord AB is at a perpendicular distance of 5 cm from the center, and chord CD is at a perpendicular distance of 12 cm from the center. Calculate the lengths of both chords, and state which chord is longer and why.
Solutions and Explanations
Solution 1: Let chord AB = 8 cm. The perpendicular from center O bisects AB, so half-chord is 8 / 2 = 4 cm. Triangle with radius r = 5 cm and half-chord 4 cm is a right triangle. By the Pythagorean theorem, distance d = sqrt(5^2 - 4^2) = sqrt(25 - 16) = sqrt(9) = 3 cm. Similarly, chord CD = 8 cm has half-chord 4 cm, so its distance from the center is also sqrt(5^2 - 4^2) = 3 cm. Because both chords have the exact same distance of 3 cm from the center, they are equidistant from the center, confirming the theorem.
Solution 2: Let chords AB and CD intersect at point X. Drop perpendiculars OL to AB and OM to CD. Because chords AB and CD are equal in length (AB = CD), by the Equidistant Chords Theorem, their perpendicular distances from the center are equal: OL = OM. Now consider right triangles OLX and OMX: Angle OLX = Angle OMX = 90 degrees (by construction); hypotenuse OX is shared by both triangles (OX = OX); and leg OL = leg OM (equal chords are equidistant). By the RHS Congruence Criterion, Triangle OLX =~ Triangle OMX. By CPCTC, corresponding angles are equal: Angle OX L = Angle OXM. This proves that segment OX bisects Angle LXM (the angle between the two chords).
Solution 3: For chord AB, distance d_1 = 5 cm and radius r = 13 cm. Half-chord length is sqrt(r^2 - d_1^2) = sqrt(13^2 - 5^2) = sqrt(169 - 25) = sqrt(144) = 12 cm. The full length of chord AB is 2 * 12 = 24 cm. For chord CD, distance d_2 = 12 cm. Half-chord length is sqrt(r^2 - d_2^2) = sqrt(13^2 - 12^2) = sqrt(169 - 144) = sqrt(25) = 5 cm. The full length of chord CD is 2 * 5 = 10 cm. Chord AB (24 cm) is strictly longer than chord CD (10 cm). This illustrates the fundamental geometric principle that chords closer to the center (distance 5 cm vs 12 cm) are longer than chords farther away.