Ninth Grade Mathematics: Perpendicular from Center to Chord Theorems
The geometric relationship between the center of a circle and a straight chord is governed by perpendicular symmetry. In this chapter, you will master the foundational theorems proving that a perpendicular dropped from the center bisects any chord, explore the logical converse, and learn how to construct the unique circle passing through three non-collinear points.
The Perpendicular Bisector of a Chord
When a perpendicular line is dropped from the center of a circle to any chord, it automatically bisects the chord into two equal segments.
The Perpendicular Chord Bisection Theorem:
The perpendicular dropped from the center of a circle to a chord bisects the chord.
In Circle C(O, r):
If OM is perpendicular to chord AB, then AM = MB.
. - ~ ~ - .
. ' O ' .
/ /|\ \
/ r / | \ r \
; / |h \ ;
| / | \ |
\ *----+----* /
\ A M B /
. _ _ '
' - ~ ~ - .
AM == MB
Deductive Proof of the Theorem
We prove this theorem using the RHS Congruence Criterion:
Given: A circle with center O. A chord AB, and segment OM perpendicular to AB (OM perp AB).
To Prove: AM = MB.
Construction:
Draw the radii OA and OB to connect the center to the endpoints of the chord.
Proof:
Step 1: Consider Triangle OMA and Triangle OMB:
- Angle OMA = Angle OMB = 90 degrees (Given: OM is perpendicular to AB)
- Hypotenuse OA = Hypotenuse OB = r (Radii of the same circle)
- Side OM = Side OM (Common Leg)
Step 2: By the RHS Congruence Criterion:
Triangle OMA =~ Triangle OMB.
Step 3: By CPCTC:
AM = MB.
The perpendicular line dropped from the center always bisects the chord. Furthermore, by CPCTC, Angle AOM = Angle BOM, proving that OM also bisects the central angle subtended by the chord.
The Converse Theorem
The converse statement is equally foundational:
Converse Theorem:
The line segment joining the center of a circle to the midpoint of a chord
is strictly perpendicular to the chord.
In Circle C(O, r):
If M is the midpoint of chord AB (AM = MB), then OM is perpendicular to AB (OM perp AB).
Proof: In Triangle OMA and Triangle OMB, OA = OB (radii), AM = MB (given: M is midpoint), and OM = OM (common). By the SSS Congruence Criterion, Triangle OMA =~ Triangle OMB. By CPCTC, Angle OMA = Angle OMB. Because these adjacent angles form a linear pair along chord AB, Angle OMA + Angle OMB = 180 degrees, so 2*Angle OMA = 180, which proves Angle OMA = 90 degrees. Thus, OM is perpendicular to AB.
Right Triangle Relationship: The Radius, Chord, and Distance
Connecting the center O to an endpoint A of the chord and dropping perpendicular OM forms a right-angled triangle OMA with legs AM and OM, and hypotenuse OA = r:
The Fundamental Pythagorean Chord Formula:
r^2 = d^2 + (c / 2)^2
where:
r is the radius of the circle (OA)
d is the perpendicular distance from the center to the chord (OM)
c is the total length of the chord (AB = 2 * AM)
Knowing any two of these three geometric quantities immediately determines the third.
Three Non-Collinear Points Determine a Unique Circle
How many circles can pass through a given set of points? Through a single point: Infinitely many circles can be drawn. Through two distinct points: Infinitely many circles can be drawn (their centers all lie on the perpendicular bisector of the segment connecting the two points). Through three points:
Theorem:
There is one and only one unique circle passing through three given
non-collinear points.
To find the unique center, draw the perpendicular bisectors of the two segments connecting the three points. Because the lines are non-parallel, their perpendicular bisectors intersect at a single unique point O (the circumcenter), which is equidistant from all three points.
Chapter Exercises
Exercise 1: A chord of length 16 cm is drawn in a circle of radius 10 cm. Calculate the perpendicular distance of the chord from the center of the circle.
Exercise 2: A perpendicular dropped from the center of a circle to a chord has length 5 cm. If the radius of the circle is 13 cm, calculate the total length of the chord.
Exercise 3: Two parallel chords of lengths 12 cm and 16 cm are drawn on the same side of the center of a circle of radius 10 cm. Find the perpendicular distance between the two chords.
Solutions and Explanations
Solution 1: Let the chord be AB = 16 cm, and the radius be r = 10 cm. The perpendicular OM from center O bisects the chord, so AM = AB / 2 = 16 / 2 = 8 cm. Triangle OMA is a right-angled triangle with hypotenuse OA = r = 10 cm and leg AM = 8 cm. By the Pythagorean theorem: OM^2 + AM^2 = OA^2, so OM^2 + 8^2 = 10^2, which gives OM^2 + 64 = 100. Subtract 64 from both sides: OM^2 = 36. Taking the square root gives perpendicular distance OM = 6 cm.
Solution 2: Let the perpendicular distance from center O to chord AB be OM = 5 cm, and radius OA = 13 cm. Triangle OMA is a right-angled triangle. By the Pythagorean theorem: AM^2 + OM^2 = OA^2, so AM^2 + 5^2 = 13^2, which gives AM^2 + 25 = 169. Subtract 25: AM^2 = 144. Taking the square root gives half-chord AM = 12 cm. Since the perpendicular bisects the chord, the total length of the chord is AB = 2 * AM = 2 * 12 = 24 cm.
Solution 3: Let the two parallel chords be AB = 16 cm and CD = 12 cm, with radius r = 10 cm. Since both chords are on the same side of center O, drop a single perpendicular line from O cutting AB at M and CD at N. Since OM is perpendicular to AB, it bisects AB: AM = 16 / 2 = 8 cm. In right triangle OMA: OM = sqrt(r^2 - AM^2) = sqrt(10^2 - 8^2) = sqrt(100 - 64) = sqrt(36) = 6 cm. Since ON is perpendicular to CD, it bisects CD: CN = 12 / 2 = 6 cm. In right triangle ONC: ON = sqrt(r^2 - CN^2) = sqrt(10^2 - 6^2) = sqrt(100 - 36) = sqrt(64) = 8 cm. The perpendicular distance between the two parallel chords is the difference between their distances from the center: MN = ON - OM = 8 - 6 = 2 cm.