Circles: A Complete Guide for Class 9 Maths (NCERT)

Basic Terms Related to a Circle

Circle

A collection of all points in a plane that are at a fixed distance from a fixed point in the plane.

Centre

The fixed point from which all points on the circle are equidistant. Usually denoted by 'O'.

Radius

The fixed distance from the centre to any point on the circle. Plural is 'radii'. Denoted by 'r'.

Chord

A line segment joining any two points on a circle.

Diameter

A chord that passes through the centre of the circle. It is the longest chord and is equal to twice the radius (d = 2r).

Arc

A piece of a circle between two points. It can be a minor arc (shorter part) or a major arc (longer part).

Segment

The region between a chord and either of its arcs. A circle has a minor segment and a major segment.

Sector

The region between an arc and the two radii joining the centre to the end points of the arc. A circle has a minor sector and a major sector.

Important Circle Theorems for Class 9

Theorems are the heart of geometry. They are proven statements that we can use as rules to solve problems. For circles, there are a few foundational theorems in your Class 9 syllabus that you must know perfectly.

1. Equal Chords and Their Angles at the Centre (Theorems 10.1 & 10.2):
Think of two equal-length sticks (chords) inside a circle. If you draw lines from the ends of each stick to the center, the angles formed at the center will be equal. The reverse is also true: if the angles at the center are equal, the chords making those angles must be equal in length. This is a powerful tool for proving triangles congruent within a circle.

2. Perpendicular from the Centre to a Chord (Theorems 10.3 & 10.4):
This is one of the most frequently used theorems. If you draw a line from the centre of a circle straight down to a chord so that it forms a 90° angle, that line will cut the chord into two equal halves (it bisects the chord). This creates a right-angled triangle, allowing you to use the Pythagoras theorem to find the radius, chord length, or distance from the centre. Its converse is also true: a line from the centre that bisects a chord is perpendicular to it.

3. Angle Subtended by an Arc (Theorem 10.8):
Imagine an arc (a piece of the circle's boundary). The angle it forms at the center of the circle is always exactly double the angle it forms at any point on the remaining part of the circle's circumference. For example, if an arc makes a 100° angle at the center, it will make a 50° angle at any point on the major arc. This theorem is crucial for finding unknown angles.

Worked Example: Applying Circle Theorems

• Problem: A chord of a circle of radius 13 cm is at a distance of 5 cm from the centre. Find the length of the chord. Step 1: Draw and label a diagram. Draw a circle with centre O. Draw a chord AB. Draw a line from O perpendicular to AB, meeting AB at point M. Label the radius OA = 13 cm and the perpendicular distance OM = 5 cm. Step 2: Identify the relevant theorem. The line from the centre perpendicular to a chord bisects the chord (Theorem 10.3). This means OM ⊥ AB and AM = MB. This also means that ΔOMA is a right-angled triangle with the right angle at M. Step 3: Apply the Pythagoras Theorem. In the right-angled triangle ΔOMA: OA² = OM² + AM² 13² = 5² + AM² 169 = 25 + AM² Step 4: Solve for AM. AM² = 169 - 25 AM² = 144 AM = √144 AM = 12 cm Step 5: Calculate the full length of the chord AB. Since the perpendicular from the centre bisects the chord, AB = 2 × AM. AB = 2 × 12 cm AB = 24 cm Final Answer: The length of the chord is 24 cm.

Exam Tips & Common Mistakes

When solving problems on circles, students often make a few common errors. Keep these points in mind to secure full marks:

• State the Theorem: Always mention the full statement of the theorem you are using in your solution. For example, write, "By the theorem that the perpendicular from the centre to a chord bisects the chord, we have AM = MB." This shows the examiner you understand the logic.
• Sector vs. Segment: Do not confuse a sector with a segment. A sector is a pizza slice (formed by two radii and an arc). A segment is the area cut off by a chord (formed by a chord and an arc).
• Pythagoras Errors: Be careful with calculations when using the Pythagoras theorem. A common mistake is mixing up the hypotenuse (which is always the radius in these problems) with the other sides.
• Angle at the Centre Theorem: Remember the angle at the centre is double the angle at the circumference, not the other way around. Double-check which angle you are given and which you need to find.

Practice Questions with Solutions

• Q: The radius of a circle is 8 cm and the length of one of its chords is 12 cm. Find the distance of the chord from the centre. A: Step 1: Draw a circle with centre O, radius OA = 8 cm, and chord AB = 12 cm. Draw OM perpendicular to AB. Step 2: By the theorem that the perpendicular from the centre bisects the chord, AM = AB/2 = 12/2 = 6 cm. Step 3: In the right-angled triangle ΔOMA, apply Pythagoras theorem: OA² = OM² + AM². Step 4: Substitute the values: 8² = OM² + 6². This gives 64 = OM² + 36. Step 5: Solve for OM: OM² = 64 - 36 = 28. So, OM = √28 = 2√7 cm. Final answer: The distance of the chord from the centre is 2√7 cm.
• Q: In a circle, two equal chords AB and CD intersect at point P. Prove that the segments of one chord are equal to the corresponding segments of the other chord (AP = DP and CP = BP). A: Step 1: Given AB = CD. Draw perpendiculars OE and OF from the centre O to the chords AB and CD respectively. In ΔOEP and ΔOFP, OE = OF (equal chords are equidistant from the centre), OP = OP (common), and ∠OEP = ∠OFP = 90°. Step 2: By RHS congruence, ΔOEP ≅ ΔOFP. Therefore, PE = PF (by CPCTC). Step 3: Since AB = CD, their halves are also equal. AE = AB/2 and DF = CD/2, so AE = DF. Step 4: To prove AP = DP, we write AP = AE + EP. Since AE = DF and EP = FP, we can write AP = DF + FP = DP. Step 5: To prove CP = BP, we subtract the equal segments from the equal chords: AB - AP = CD - DP, which gives BP = CP. Final answer: Hence, it is proved that the corresponding segments are equal.
• Q: If ∠AOB = 90° and ∠BOC = 120°, where A, B, and C are points on a circle with centre O, find ∠ABC. A: Step 1: First, find the angle ∠AOC. The angles around the centre O sum to 360°. ∠AOC (reflex) = ∠AOB + ∠BOC = 90° + 120° = 210°. Therefore, ∠AOC (non-reflex) = 360° - 210° = 150°. Step 2: The angle subtended by the arc AC at the centre is ∠AOC = 150°. Step 3: The angle subtended by the arc AC at any point on the remaining part of the circle (like point B) is half the angle at the centre. This is a property based on Theorem 10.8. Step 4: Therefore, ∠ABC = (1/2) ∠AOC = (1/2) 150° = 75°. Final answer: ∠ABC = 75°.
• Q: A circle has a radius of 5 cm. A chord of this circle is 8 cm long. How many such chords can be drawn? A: Step 1: We can find the distance of this chord from the centre. Let the radius be r=5 and chord length be L=8. Let the distance be d. We know r² = d² + (L/2)². Step 2: Substitute values: 5² = d² + (8/2)². So, 25 = d² + 4². This gives 25 = d² + 16. Step 3: Solve for d: d² = 25 - 16 = 9. So, d = 3 cm. Step 4: All chords that are at a distance of 3 cm from the centre will have a length of 8 cm. Since we can draw a circle of radius 3 cm inside the larger circle, there are infinitely many points on this inner circle. We can draw a tangent from each of these points to form a chord of the outer circle. Therefore, there are infinitely many such chords. Final answer: Infinitely many chords of length 8 cm can be drawn.

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