Welcome to our deep dive into CBSE Class 10 Maths Chapter 10: Circles. This is more than a summary; it’s a complete pedagogical analysis of the chapter’s design and strategy. We deconstruct the core proofs for Theorem 10.1 and Theorem 10.2 and provide a thematic analysis of all problem archetypes in Exercise 10.2—from “Circumscribing Polygons” to the 5-mark “Pinnacle Problems.” Explore the logic, understand the exam weightage, and interact with a flowchart that maps out the problem-solving process.
This chapter has two main ideas. First, a tangent and a radius are perpendicular. Second, two tangents from one external point are equal in length. This page explains these theorems and shows how to solve problems with them.
A line in a plane can have one of three relationships with a circle.
The line does not touch the circle at all. (0 common points)
The line cuts through the circle at two distinct points.
The line touches the circle at exactly one point. This point is the ‘point of contact’.
The number of tangents you can draw to a circle depends on where the point is located.
Any line through P will cut the circle at two points (a secant).
Result: 0 Tangents
There is only one line that touches the circle at this single point.
Result: 1 Tangent
There are exactly two lines from P that touch the circle. (Theorem 10.2).
Result: 2 Tangents
“The tangent at any point of a circle is perpendicular to the radius through the point of contact.”
This means if XY is a tangent at point P, and O is the center, the radius OP will form a 90-degree angle with the line XY. So, OP is perpendicular to XY.
“The lengths of tangents drawn from an external point to a circle are equal.”
If you pick a point P outside the circle and draw two tangents that touch the circle at points Q and R, then the lengths PQ and PR will be equal.
This RHS proof is powerful because it gives two results for the price of one (from CPCT – Corresponding Parts of Congruent Triangles):
| Feature | Theorem 10.1 (Tangent-Radius) | Theorem 10.2 (External Point) |
|---|---|---|
| Main Idea | A tangent and radius are perpendicular (form a 90° angle) at the point of contact. | Tangents from a single external point are equal in length. |
| Number of Tangents | Deals with one tangent at a time. | Deals with a pair of tangents from one point. |
| Proof Method | Indirect proof (using “shortest distance”). | Direct proof (using RHS triangle congruence). |
| Key Result | angle = 90° | Side PQ = Side PR |
| Common Use | To create right-angled triangles for using the Pythagorean theorem. | To set up algebraic equations (e.g., x+8 = 15) and for proofs involving isosceles triangles. |
Drag the external point P (the blue dot) with your mouse or finger. Observe how the two tangent lengths, PA and PB, change but always remain equal to each other.
This demo uses geometry to find the tangent points from point P.
The problems in the textbook are not random. They test specific applications of the theorems. Use the filters to see examples of each type.
These are direct applications. Q2 and Q10 (proof) show that the angle at the center and the angle between tangents are supplementary (add to 180°). Q7 is a numerical version of Example 1, combining Thm 10.1 with Class 9 chord properties.
These problems ask you to prove corollaries. Q4 (tangents at ends of a diameter are parallel) is a key result of Thm 10.1. Q9 (Prove angle AOB = 90°) is a clever proof using SSS congruence from Thm 10.2.
This family of problems uses one strategy: apply Thm 10.2 to every vertex, then add the equations. Q8 (AB+CD = AD+BC) leads directly to Q11 (a circumscribing parallelogram must be a rhombus).
This is the capstone problem. It requires you to find the area of the triangle in two different ways (Heron’s Formula, and 1/2 × base × height) and set them equal. It combines Thm 10.2 with complex algebra.
Not sure which theorem to use? Follow this logic flow to decide how to approach a problem. This chart shows the general decision-making process.
This flowchart shows a simplified path. Some problems (like Q12) combine multiple methods.
Let’s deconstruct the logic for two common exam proofs from Exercise 10.2.
Given: Two parallel tangents XY and X’Y’, and a third tangent AB at point C.
Logic Step 1: Look at the external point A. Tangents AP and AC are drawn.
Result (Thm 10.2): Triangle OPA is congruent to Triangle OCA (by SSS or RHS).
Key Takeaway: angle POA = angle COA. Let’s call this ‘x’. So angle POC = 2x.
Logic Step 2: Look at the external point B. Tangents BQ and BC are drawn.
Result (Thm 10.2): Triangle OQB is congruent to Triangle OCB (by SSS or RHS).
Key Takeaway: angle QOB = angle COB. Let’s call this ‘y’. So angle QOC = 2y.
Logic Step 3: Line POQ is a straight line (diameter) because the tangents are parallel.
Result: The angle on the straight line is 180°.
Equation: angle POC + angle QOC = 180°
Final Step: Substitute from Steps 1 and 2.
(2x) + (2y) = 180° -> 2(x + y) = 180° -> x + y = 90°
Conclusion: Since angle AOB = angle COA + angle COB = x + y, we have angle AOB = 90°.
Given: A quadrilateral ABCD circumscribing a circle (center O).
Logic Step 1: This is the same logic as Q9, but applied four times.
At Vertex A: angle 1 = angle 2
At Vertex B: angle 3 = angle 4
At Vertex C: angle 5 = angle 6
At Vertex D: angle 7 = angle 8
Logic Step 2: The sum of all angles around the center point O is 360°.
Equation: angle 1 + angle 2 + angle 3 + angle 4 + angle 5 + angle 6 + angle 7 + angle 8 = 360°
Logic Step 3: Substitute using the pairs from Step 1.
( angle 2 + angle 2 ) + ( angle 3 + angle 3 ) + ( angle 6 + angle 6 ) + ( angle 7 + angle 7 ) = 360°
Simplified: 2(angle 2 + angle 3 + angle 6 + angle 7) = 360°
Result: angle 2 + angle 3 + angle 6 + angle 7 = 180°
Final Step: Group the angles by the opposite sides.
Angle for side AB: angle AOB = angle 2 + angle 3
Angle for side CD: angle COD = angle 6 + angle 7
Conclusion: (angle 2 + angle 3) + (angle 6 + angle 7) = 180° -> angle AOB + angle COD = 180°.
In the study of Euclidean geometry, the relationship between a line and a circle stands as a cornerstone of geometric inquiry. Understanding the ways in which these fundamental figures can interact in a plane provides the basis for numerous theorems and applications. Of the possible configurations, the case of tangency—where a line touches a circle at a single point—is of particular importance, revealing deep and elegant properties of circular forms.
To begin our exploration, we must first establish a formal definition of a circle. A circle is the set of all points in a plane that are at a constant distance, known as the radius, from a fixed point, the centre. Given this definition, we can classify the three possible spatial relationships between a line and a circle within a shared plane:
The objective of this monograph is to systematically explore the properties, foundational theorems, and geometric propositions related to the tangent line, providing a comprehensive treatise on its significance.
To fully appreciate the unique characteristics of a tangent, it is strategically important to view it not merely as a static definition but as a dynamic, limiting case of a secant. This perspective offers a more profound insight into its geometric properties and its relationship to the circle it touches.
The concept of a tangent can be deconstructed as a special case of a secant. Consider a secant line intersecting a circle at two points, P and Q. If we fix point P and rotate the secant line around it, the second point of intersection, Q, will move along the circumference of the circle. As the line rotates, Q will approach P. The limiting position of the secant, at the exact moment when point Q coincides with P, is the tangent line at point P. In this view, the tangent is the result of a secant whose corresponding chord length has diminished to zero.
This same principle can be observed from a different perspective by considering a series of parallel secants. If one draws a secant and then a series of lines parallel to it, the chords formed by the intersection of these lines with the circle will gradually decrease in length. Eventually, there will be two positions, one on each side of the original secant, where the length of the chord becomes zero. At these positions, the secant lines become tangent to the circle.
From this analysis, we can derive several key definitions and principles:
Having established a robust conceptual definition of the tangent, we now proceed to a rigorous examination of the foundational theorems that govern its behavior.
The true power of geometric analysis lies in its ability to formalize observed relationships into rigorously proven theorems. These theorems provide an unshakeable foundation upon which more complex propositions can be built. This section will present and prove the two central theorems that define the essential properties of tangents to a circle.
This first theorem establishes the fundamental relationship between a tangent and the radius drawn to its point of contact.
Theorem 10.1: The tangent at any point of a circle is perpendicular to the radius through the point of contact.
Proof:
From this seminal theorem, two important corollaries arise:
This second theorem addresses the case of two tangents drawn from a single external point and proves a crucial property regarding their lengths. First, we must define the term length of the tangent as the length of the line segment from the external point to the point of contact on the circle.
Theorem 10.2: The lengths of tangents drawn from an external point to a circle are equal.
Proof using Right-Hand-Side (RHS) Congruence:
An Alternate Proof via Pythagorean Theorem
The same result can be derived using the Pythagorean Theorem. In the right triangle Triangle OQP: PQ2 = OP2 – OQ2
And in the right triangle Triangle ORP: PR2 = OP2 – OR2
Since OQ = OR (radii of the same circle), it follows that: PQ2 = PR2, which implies PQ = PR.
A further property can be concluded from the congruence of Triangle OQP and Triangle ORP: because angle OPQ = angle OPR, the line segment OP is the angle bisector of angle QPR. This means the centre of the circle always lies on the bisector of the angle formed between the two tangents.
The true power and elegance of geometric theorems are revealed when they are applied to solve specific problems or establish new propositions. The core theorems governing tangents serve as foundational axioms from which a host of other geometric truths can be derived. This section reframes several illustrative examples as formal propositions to demonstrate the practical utility of these theorems.
Proposition: In two concentric circles, the chord of the larger circle which is tangent to the smaller circle is bisected at the point of contact.
Proof: Let C₁ be the larger circle and C₂ be the smaller circle, with a common centre O. Let AB be a chord of C₁ that is tangent to C₂ at point P. By joining O to P, we form the radius OP of the smaller circle C₂.
According to Theorem 10.1, the radius to the point of contact is perpendicular to the tangent. Therefore, OP is perpendicular to AB.
Now, considering the larger circle C₁, AB is a chord and OP is a line segment from the centre that is perpendicular to this chord. A fundamental property of circles states that a perpendicular drawn from the centre to a chord bisects the chord.
Therefore, AP = BP, and the chord AB is bisected at the point of contact P.
Proposition: The angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
Proof: Let P be an external point from which two tangents, PQ and PR, are drawn to a circle with centre O. Here, Q and R are the points of contact. The angle between the tangents is angle QPR, and the angle subtended by the line segment QR at the centre is angle QOR.
By Theorem 10.1, the radii OQ and OR are perpendicular to the tangents PQ and PR, respectively. Thus, angle OQP = 90° and angle ORP = 90°.
The figure OQPR is a quadrilateral. The sum of the interior angles of a quadrilateral is 360°. Therefore: angle OQP + angle QPR + angle ORP + angle QOR = 360°
Substituting the known values of the right angles: 90° + angle QPR + 90° + angle QOR = 360° which simplifies to 180° + angle QPR + angle QOR = 360°
Solving for the sum of the remaining angles gives: angle QPR + angle QOR = 180°
Thus, the angle between the tangents and the angle at the centre are supplementary.
Proposition: The tangents drawn at the ends of a diameter of a circle are parallel.
Proof: Let AB be a diameter of a circle with centre O. Let L₁ be the tangent to the circle at point A and L₂ be the tangent at point B. The diameter AB can be viewed as a straight line passing through the centre. The segments OA and OB are radii.
According to Theorem 10.1, the tangent is perpendicular to the radius at the point of contact. Therefore, L₁ is perpendicular to AB at point A, and L₂ is perpendicular to AB at point B.
Since both lines L₁ and L₂ are perpendicular to the same line AB, they must be parallel to each other. These propositions are direct and powerful extensions of the core theorems, illustrating how foundational principles are applied to deduce more complex geometric relationships.
This monograph has systematically explored the geometric properties of a tangent to a circle, from its conceptual definition to its core theorems and applications. The principal findings presented herein are consolidated below for clarity and retention.
Yes. It is a high-weightage chapter. In the 2024-2025 board exam structure, the “Geometry” unit (which includes Circles and Triangles) has 22 marks. This single “Circles” chapter is often allotted around 8 marks, which may include MCQs and one 5-mark long-answer question.
Yes, you need to know one theorem from Class 9: “The perpendicular from the centre of a circle to a chord bisects the chord.” This is used in Example 1 and Exercise 10.2, Q7 (the concentric circles problem).
Many students find Exercise 10.2, Q12 to be the most difficult. This is the problem where a triangle circumscribes a circle of radius 4 cm, and you have to find the sides AB and AC. It requires using Theorem 10.2 (AE = AF = x) and then finding the triangle’s area by two different methods (Heron’s formula and 1/2 × base × height) and setting them equal.
