Circles Class 10 Maths Chapter Notes

Key Theorems and Properties

• A circle can have infinitely many tangents.
• A tangent to a circle intersects it at exactly one point, known as the point of contact.
• The tangent at any point of a circle is perpendicular to the radius through the point of contact. (Theorem 10.1)
• A line drawn through the end of a radius and perpendicular to it is a tangent to the circle.
• The lengths of tangents drawn from an external point to a circle are equal. (Theorem 10.2)
• From a point inside a circle, no tangents can be drawn.
• From a point on the circle, exactly one tangent can be drawn.
• From a point outside the circle, exactly two tangents can be drawn.
• If two tangents are drawn from an external point, they subtend equal angles at the center.
• Tangents drawn from an external point are equally inclined to the line segment joining the center to that point.

Glossary of Terms

Circle

A collection of all points in a plane that are at a fixed distance (radius) from a fixed point (center).

Tangent

A line that intersects the circle at exactly one point.

Secant

A line that intersects the circle at two distinct points.

Point of Contact

The single point where a tangent touches the circle.

Radius

The line segment from the center of a circle to any point on the circle. It is also the length of this segment.

Chord

A line segment whose endpoints both lie on the circle.

Concentric Circles

Circles with the same center but different radii.

Theorem 10.2: Tangents from an External Point

One of the most important theorems in this chapter states that the lengths of tangents drawn from an external point to a circle are equal. Let's understand the proof, as it's frequently asked in exams.

Consider a circle with center O. Let P be an external point from which two tangents, PA and PB, are drawn to the circle, touching it at points A and B respectively. We need to prove that PA = PB.

To prove this, we join OA, OB, and OP.

Now, consider the two triangles, ΔOAP and ΔOBP.

1. OA = OB: Both are radii of the same circle.
2. OP = OP: This is the common side for both triangles.
3. ∠OAP = ∠OBP = 90°: This is because the tangent at any point is perpendicular to the radius at the point of contact (Theorem 10.1).

Since both triangles are right-angled, and we have the hypotenuse and one side equal, we can prove their congruence using the RHS (Right angle-Hypotenuse-Side) congruence rule.

Therefore, ΔOAP ≅ ΔOBP.

By CPCTC (Corresponding Parts of Congruent Triangles), the corresponding sides must be equal. Hence, PA = PB. This proves the theorem. Additionally, by CPCTC, we also get ∠AOP = ∠BOP and ∠APO = ∠BPO, which means the tangents are equally inclined to the line segment OP.

Solved Examples

• {"title":"Example 1: Finding Tangent Length","bodyMarkdown":"Problem: A tangent PQ at a point P of a circle of radius 5 cm meets a line through the center O at a point Q so that OQ = 13 cm. Find the length of PQ.\n\nSolution:\nGiven: Radius OP = 5 cm, OQ = 13 cm.\nWe know that the tangent is perpendicular to the radius at the point of contact (∠OPQ = 90°).\nSo, ΔOPQ is a right-angled triangle.\nUsing Pythagoras theorem: OQ² = OP² + PQ²\n13² = 5² + PQ²\n169 = 25 + PQ²\nPQ² = 169 - 25 = 144\nPQ = √144 = 12 cm.\nAnswer: The length of the tangent PQ is 12 cm."}
• {"title":"Example 2: Angle Property","bodyMarkdown":"Problem: Two tangents TP and TQ are drawn to a circle with center O from an external point T. Prove that ∠PTQ = 2∠OPQ.\n\nSolution:\nLet ∠PTQ = θ.\nSince tangents from an external point are equal, TP = TQ. So, ΔTPQ is an isosceles triangle.\n∠TPQ = ∠TQP = (180° - θ) / 2 = 90° - θ/2.\nWe know ∠OPT = 90° (tangent is ⊥ to radius).\nSo, ∠OPQ = ∠OPT - ∠TPQ\n∠OPQ = 90° - (90° - θ/2)\n∠OPQ = θ/2\nThis implies θ = 2∠OPQ, which means ∠PTQ = 2∠OPQ. Hence proved."}

Construction: Tangents from an External Point

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Exam Tips and Common Mistakes

• Don't confuse secant and tangent. A tangent touches at one point; a secant intersects at two. Read the question carefully.
• Remember the two main theorems. Most questions are direct applications of Theorem 10.1 (Radius ⊥ Tangent) and Theorem 10.2 (Equal tangents from an external point).
• Practice proofs. The proof for Theorem 10.2 is very common. Make sure you can write it step-by-step with correct reasons (like RHS congruence).
• Use Pythagoras Theorem correctly. Many problems combine tangents with right-angled triangles. Be careful with calculations.
• Draw neat diagrams. A clear, well-labeled diagram can help you visualize the problem and often earns you marks, even if the final calculation is incorrect.

Practice Questions with Solutions

• How many tangents can be drawn to a circle parallel to a given secant? Two. One on each side of the secant.
• What is the angle between a tangent and the radius at the point of contact? 90 degrees.
• The lengths of two tangents from an external point P to a circle are 8 cm. What is the relation between them? They are equal. This is a direct application of Theorem 10.2.
• A point P is 10 cm from the center of a circle. The length of the tangent drawn from P to the circle is 8 cm. What is the radius of the circle? Using Pythagoras theorem, r² + 8² = 10². So, r² = 100 - 64 = 36. The radius is 6 cm.

Frequently Asked Questions