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Geometry

Arc and chord of circle

The geometry of arcs and chords includes perpendicular bisectors, subtended angles, cyclic properties, and chord lengths.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade
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Definition and statement

Perpendicular Bisector and Equal Chords Theorem
The perpendicular from the centre of a circle to a chord bisects the chord; conversely, the line drawn through the centre of a circle to bisect a chord is perpendicular to the chord. Furthermore, equal chords are equidistant from the centre and subtend equal angles at the centre.

📐 Geometric Interpretation

A chord divides a circle into two arcs (major and minor). Drawing a perpendicular from the centre creates two congruent right-angled triangles via the Radius-Hypotenuse-Side (RHS) congruence criterion, visually establishing the bisecting property.

⚙️ Algebraic & Structural Meaning

Translates Euclidean geometric constraints into coordinate geometry equations using distance formulas, circle equations $(x-h)^2 + (y-k)^2 = r^2$, and line-intersection parameters.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition Domain / Interval Necessity What Fails If Condition Dropped
Circle Centred at Origin/Fixed Point Metric Space / Euclidean Plane Strictly Mandatory / Necessary Without a fixed centre, the concepts of equidistant chords and angular subtractions lose their reference frame.
Chord Length $L \le 2r$ (Diameter) Length domain $(0, 2r]$ Strictly Mandatory If the specified chord length exceeds the diameter $2r$, no real Euclidean chord can exist within the circle.
Perpendicularity Condition Orthogonality ($\theta = 90^\circ$) Essential for bisection property A non-perpendicular line passing through the centre and intersecting a chord will divide the chord into unequal segments.

Useful results and shortcuts

Use these identities and results when their conditions apply:

01

Chord Length from Radius and Perpendicular Distance

$$ L = 2 \sqrt{r^2 - d^2} $$

Rule: Given radius $r$ and perpendicular distance $d$ from the centre to the chord, directly compute the chord length $L$ using Pythagoras theorem.

Application: Frequently asked in UP TGT/PGT where numerical values for $r$ and $d$ are given to find the length of the chord.
02

Intersecting Chords Theorem (Internal & External)

$$ PA \cdot PB = PC \cdot PD $$

Rule: If two chords $AB$ and $CD$ intersect inside or outside a circle at point $P$, the products of their segment lengths are equal.

03

Perpendicular Distance Between Parallel Chords

$$ d = \sqrt{r^2 - \left(\frac{L_1}{2}\right)^2} \pm \sqrt{r^2 - \left(\frac{L_2}{2}\right)^2} $$

Rule: For parallel chords on the same side of the centre, use the difference of distances; for opposite sides, use the sum.

04

Angle Subtended by Arc at Centre vs Circumference

$$ \theta_{\text{centre}} = 2 \theta_{\text{circumference}} $$

Rule: The angle subtended by an arc at the centre is double the angle subtended by the same arc at any point on the remaining circumference.

Application: Used in combination with cyclic quadrilateral properties in geometry problem-solving.
05

Length of Common Chord of Two Intersecting Circles

$$ L = \frac{2}{d} \sqrt{s(s-r_1)(s-r_2)(s-d)} $$

Rule: Using Heron's formula for the triangle formed by centres and intersection points, where $s = \frac{r_1 + r_2 + d}{2}$ and $d$ is distance between centres.

Application: Advanced coordinate geometry and circle combination problems in LT Grade and UP PGT.
06

Rapid Elimination via Triangle Inequality

$$ 0 < L \le 2r $$

Rule: Instantly eliminate options where chord length exceeds the diameter or where segment sums violate triangle inequalities.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Confusing Same Side vs Opposite Side of Centre for Parallel Chords

Common Pitfall: Candidates always add or always subtract the perpendicular distances of parallel chords without checking if they lie on the same side or opposite sides of the centre.

Counterexample: Chords of lengths $6$ cm and $8$ cm in a circle of radius $5$ cm.

Why it fails: Distances are $4$ cm and $3$ cm. Distance between them is $4 - 3 = 1$ cm (same side) or $4 + 3 = 7$ cm (opposite sides). Choosing the wrong configuration leads to a wrong option.

🛡️ Exam Defense: Carefully read whether chords are on 'the same side' or 'opposite sides' of the centre.
⚠️ COMMON ERROR #02

Assuming Any Line Through Centre Bisects a Chord

Common Pitfall: Assuming a diameter passing through a chord always bisects it regardless of perpendicularity.

Counterexample: A non-perpendicular secant line passing through the centre.

Why it fails: Only the perpendicular dropped from the centre bisects the chord.

🛡️ Exam Defense: Verify the $90^\circ$ orthogonality condition before applying bisection formulas.

Comparison

Concept / TheoremHypotheses / AssumptionsKey Formula / OutcomeGeometric MeaningExam Identification Clue
Perpendicular BisectorLine from centre to chord with $90^\circ$$AC = CB = \frac{L}{2}$Divides chord and arc into two equal halvesKeywords: 'perpendicular', 'bisects chord'
Equal Chords & DistanceChords of equal length in circle$d_1 = d_2$Equidistant chords are congruent in subtended propertiesKeywords: 'equidistant', 'equal chords'
Intersecting ChordsChords intersect at internal/external point $P$$PA \cdot PB = PC \cdot PD$Power of a point theorem applicationKeywords: 'intersecting chords', 'secants'
Angle at CentreArc subtending angles at centre and circumference$\angle O = 2 \angle C$Central angle is twice the inscribed angleKeywords: 'subtends', 'centre', 'circumference'

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the length of a chord which is at a distance of $4$ cm from the centre of a circle of radius $5$ cm.

(A) $6$ cm(B) $8$ cm(C) $10$ cm(D) $3$ cm
✓ Correct Answer: (B)
📚 Method A: Formal Textbook Derivation
Step 1: Identify $r = 5$, $d = 4$. Step 2: Use half-chord relation $x = \sqrt{r^2 - d^2} = \sqrt{5^2 - 4^2} = \sqrt{9} = 3$. Step 3: Total chord length $L = 2x = 6$? Wait, $\sqrt{25-16} = 3$, so $2(3) = 6$. Let's check options: (A) $6$ cm.
⚡ Method B: 25-Second Exam Shortcut
Pythagorean triple $(3, 4, 5)$. Half chord is $3$, full chord is $2 \times 3 = 6$ cm. Option (A) is correct.
💡 Tip: Always remember to multiply the half-chord length by $2$.
CANONICAL PROBLEM 02 Dual-Method Breakdown

Two parallel chords of a circle of radius $10$ cm are of lengths $16$ cm and $12$ cm and lie on the same side of the centre. The distance between them is:

(A) $2$ cm(B) $4$ cm(C) $6$ cm(D) $14$ cm
✓ Correct Answer: (A)
📚 Method A: Formal Textbook Derivation
Step 1: Distance of first chord $d_1 = \sqrt{10^2 - 8^2} = 6$ cm. Step 2: Distance of second chord $d_2 = \sqrt{10^2 - 6^2} = 8$ cm. Step 3: Distance between them on the same side = $|d_2 - d_1| = |8 - 6| = 2$ cm.
⚡ Method B: 25-Second Exam Shortcut
Use Pythagorean triples $(6, 8, 10)$. Distances from centre are $6$ and $8$. Difference is $2$.
💡 Tip: Check same side vs opposite side carefully. Same side means difference; opposite side means sum ($6 + 8 = 14$).

Practice questions

Choose an answer to check it and see the solution:

Q1
MCQ 1: If the diameter of a circle is $26$ cm and the length of a chord is $24$ cm, then the distance of the chord from the centre is:
A $5$ cm
B $10$ cm
C $12$ cm
D $13$ cm
Derivation & Analysis:

Radius $r = 13$ cm, half chord $x = 12$ cm. Distance $d = \sqrt{r^2 - x^2} = \sqrt{13^2 - 12^2} = \sqrt{25} = 5$ cm. Correct option is (A).

Q2
MCQ 2: The length of a chord of a circle of radius $10$ cm subtending an angle of $90^\circ$ at the centre is:
A $5\sqrt{2}$ cm
B $10\sqrt{2}$ cm
C $10$ cm
D $20$ cm
Derivation & Analysis:

In an isosceles triangle with sides $r, r$ and angle $90^\circ$, the hypotenuse (chord length) is $r\sqrt{2} = 10\sqrt{2}$ cm. Correct option is (B).

Q3
MCQ 3: If two equal chords of a circle intersect within the circle, then the segments of one chord are respectively equal to:
A the segments of the other chord
B twice the segments of the other chord
C half the segments of the other chord
D none of these
Derivation & Analysis:

By symmetry and congruence of triangles formed by joining the centre to the intersection point and chord ends, the corresponding segments are equal.

Q4
MCQ 4: The perpendicular distance from the centre of a circle to a chord of length $16$ cm in a circle of radius $10$ cm is:
A $4$ cm
B $6$ cm
C $8$ cm
D $3$ cm
Derivation & Analysis:

$d = \sqrt{10^2 - 8^2} = \sqrt{100 - 64} = \sqrt{36} = 6$ cm. Correct option is (B).

Q5
MCQ 5: What is the maximum possible length of a chord in a circle of radius $r$?
A $r$
B $\sqrt{2}r$
C $2r$
D $\pi r$
Derivation & Analysis:

The longest chord of a circle is its diameter, which equals $2r$. Correct option is (C).

Q6
MCQ 6: If chords $AB$ and $CD$ of a circle intersect at right angles at point $P$, and $PA = 4$, $PB = 6$, $PC = 3$, then $PD$ is equal to:
A $8$
B $6$
C $4$
D $2$
Derivation & Analysis:

By intersecting chords theorem, $PA \cdot PB = PC \cdot PD \implies 4 \cdot 6 = 3 \cdot PD \implies PD = 24 / 3 = 8$. Correct option is (A).

Q7
MCQ 7: The angle subtended by a minor arc at the centre is $120^\circ$. What is the angle subtended by the corresponding major arc at the centre?
A $120^\circ$
B $240^\circ$
C $180^\circ$
D $60^\circ$
Derivation & Analysis:

The total angle around the centre is $360^\circ$. Major arc angle = $360^\circ - 120^\circ = 240^\circ$. Correct option is (B).

Q8
MCQ 8: Assertion (A): Equal chords subtend equal angles at the centre. Reason (R): Triangles formed by radii and chords are congruent by SSS criterion.
A Both (A) and (R) are true and (R) is the correct explanation of (A)
B Both (A) and (R) are true but (R) is not the correct explanation of (A)
C (A) is true but (R) is false
D (A) is false but (R) is true
Derivation & Analysis:

Both statements are mathematically rigorous and (R) correctly invokes SSS congruence ($r = r, r = r, L_1 = L_2$).

Q9
MCQ 9: If the distance of a chord from the centre is zero, the chord is:
A a tangent
B a diameter
C a point
D impossible
Derivation & Analysis:

A chord passing through the centre has zero perpendicular distance and is the diameter of the circle.

Q10
MCQ 10: Two parallel chords of length $24$ cm and $10$ cm are on opposite sides of the centre of a circle of radius $13$ cm. The distance between them is:
A $5$ cm
B $12$ cm
C $17$ cm
D $7$ cm
Derivation & Analysis:

Distances from centre are $d_1 = \sqrt{13^2 - 12^2} = 5$ cm and $d_2 = \sqrt{13^2 - 5^2} = 12$ cm. Since they are on opposite sides, total distance = $5 + 12 = 17$ cm. Correct option is (C).

Common questions

Because it converts circle geometry problems into right-angled triangles, allowing direct application of the Pythagorean theorem and trigonometric ratios.

Always draw a rough sketch. If chords are on the same side of the centre, subtract their distances from the centre. If on opposite sides, add them.

Yes, they are limiting cases and extensions of the intersecting chords theorem using the power of a point theorem.

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