A secant intersects the circle at two distinct points A and B
Chapter 10 · CBSE Class X
📘 Definition
Secant of a Circle
🎨 SVG Diagram
Secant Image
💡 Concept
Core Concepts
🔢 Formula
Important Theorem (Secant-Secant Power Theorem)
✏️ Example
Example
From an external point \(P\), two secants intersect a circle at points \(A, B\) and \(C, D\).
If \(PA = 4\text{ cm}\), \(PB = 10\text{ cm}\), and \(PC = 5\text{ cm}\), find \(PD\).
Concept Used: Secant-Secant Theorem
- Apply theorem relation
- Substitute known values
- Solve for unknown
Step by step Solution
- \[PA \cdot PB = PC \cdot PD\]
- \[4 \times 10 = 5 \times PD\]
- \[40 = 5PD\]
- \[PD = 8 \text{ cm}\]
📐 Derivation
Derivation Insight
The secant-secant theorem is derived using similar triangles formed when two secants intersect a circle. By applying AA similarity and proportionality of sides, the product relation emerges.
⚡ Exam Tip
Exam Tips for CBSE Board
⚠️ Warning
Common Mistakes
📋 Case Study
CBSE Case Study (HOTS)
A security camera scans circular regions. Two laser beams act as secants from a fixed point. If one beam length increases, analyze how the second must change to maintain constant product.
This models the secant theorem and tests conceptual clarity rather than direct computation.
📘 Definition
Definition
🎨 SVG Diagram
Tangent
Radius \(OP\) is perpendicular to the tangent at point \(P\)
📌 Note
Core Properties of a Tangent
- A tangent touches the circle at exactly one point.
- The radius drawn to the point of contact is always perpendicular to the tangent.
- No part of a tangent lies inside the circle.
- From an external point, exactly two tangents can be drawn to a circle.
- Lengths of tangents drawn from an external point are equal.
📖 Theory
Important Theorems
1. Radius-Tangent Theorem:
\[ OP \perp \text{Tangent at } P \]
2. Tangent Length Theorem:
If two tangents \(PA\) and \(PB\) are drawn from an external point \(P\), then:
\[ PA = PB \]
✏️ Example
Example
From a point \(P\), two tangents \(PA\) and \(PB\) are drawn to a circle.
If \(PA = 7\text{ cm}\), find the length of \(PB\)
Equal tangents from an external point
- Identify theorem
- Apply equality relation
- \[PA = PB\]
- \[PB = 7 \text{ cm}\]
📐 Derivation
Derivation Insight
The equality of tangents is derived using congruency of triangles. Consider triangles formed by joining the centre to the points of contact:
- \(OA = OB\) (radii)
- \(OP\) is common
- \(\angle OAP = \angle OBP = 90^\circ\)
Hence, triangles are congruent (RHS), leading to: \[ PA = PB \]
🔢 Formula
Formula Summary
⚡ Exam Tip
Exam Tips for CBSE Board
⚠️ Warning
Common Mistakes
📋 Case Study
CBSE Case Study (HOTS)
A road just touches a circular park at one point. Two observers stand at a point outside the park and measure distances along the road to two points of contact. Explain why these distances must be equal.
This real-life model applies the concept of equal tangents and tests reasoning ability.
🔬 Proof
Theorem 1: Tangent is Perpendicular to Radius
OP is perpendicular to the tangent at P
The tangent at any point of a circle is perpendicular to the radius through the point of contact.
Given
A circle with centre \(O\).
A tangent \(XY\) touches the circle at point \(P\).
TO Prove
\[OP \perp XY\]
Construction
Take any point \(Q\) on the tangent \(XY\), such that \(Q \ne P\). Join \(OQ\).
Proof
Since \(XY\) is a tangent at \(P\), it touches the circle at only one point. Therefore, any other point \(Q\) on the tangent lies outside the circle.
Hence, \[ OQ > OP \] because \(OP\) is the radius and \(OQ\) extends beyond the circle.
Now observe:
- \(OP\) is the distance from centre \(O\) to the line \(XY\)
- \(OQ\) is any other oblique distance
Therefore, \(OP\) is the shortest distance from point \(O\) to the line \(XY\).
The shortest distance from a point to a line is always the perpendicular.
Hence, \(OP\) must be perpendicular to the tangent \(XY\).
\[ OP \perp XY \]
📐 Derivation
Conceptual Insight
This theorem is fundamentally based on the minimum distance principle. Among all segments from a point to a line, the perpendicular segment is always the shortest. Since the radius to the point of contact is the shortest segment, it must be perpendicular.
✏️ Example
In a circle with centre \(O\), tangent \(AB\) touches the circle at \(P\).
If \(OP = 5\text{ cm}\), find the angle between \(OP\) and the tangent.
By theorem, \[ OP \perp AB \] \[ \angle OPB = 90^\circ \]
⚡ Exam Tip
Exam Tips
⚠️ Warning
Common Mistakes
📋 Case Study
CBSE Case Study (HOTS)
A wheel touches the ground at one point. Explain why the radius at that point is perpendicular to the ground using circle geometry.
🌟 Importance
This Theorem is Crucial for CBSE
📘 Definition
Concept Overview
🔢 Formula
Mathematical Condition
🎨 SVG Diagram
Image
Different positions of a point relative to a circle
📌 Note
Cases
- Case 1: Point Inside the Circle (\(OP < r\))
No tangent can be drawn because any line through the point will intersect the circle at two points. - Case 2: Point On the Circle (\(OP = r\))
Exactly one tangent exists at that point. It is perpendicular to the radius. - Case 3: Point Outside the Circle (\(OP > r\))
Exactly two tangents can be drawn. These tangents are equal in length.
💡 Concept
Why These Cases Occur
- Inside: Every line cuts the circle → becomes a secant → no tangent possible
- On circle: Only one line satisfies perpendicular condition → unique tangent
- Outside: Two distinct lines satisfy tangency condition → two tangents
✏️ Example
A point lies 13 cm from the centre of a circle of radius 5 cm.
How many tangents can be drawn from the point?
Compare \(OP\) with radiu
- \[OP = 13, \quad r = 5\]
- \[OP \gt r\]
- Therefore, the point lies outside the circle.
- \[\text{Number of tangents} = 2\]
📐 Derivation
Theoretical Insight
The number of tangents depends on solving the geometric condition of perpendicular distance from centre to line = radius. This condition leads to:
- No solution → no tangent
- One solution → one tangent
- Two solutions → two tangents
⚡ Exam Tip
Exam Tips
⚠️ Warning
Common Mistakes
📋 Case Study
CBSE Case Study (HOTS)
A drone is positioned relative to a circular radar region. Depending on its position, determine how many signal lines (tangents) can just touch the boundary without entering it.
🌟 Importance
Importance for Board Exams
🔬 Proof
Theorem 2: Lengths of Tangents from an External Point are Equal
Different positions of a point relative to a circle
The lengths of tangents drawn from an external point to a circle are equal.
Given
A circle with centre \(O\).
A point \(P\) lies outside the circle.
Two tangents \(PQ\) and \(PR\) are drawn to the circle touching at \(Q\) and \(R\).
To Prove
\[PQ = PR\]
Construction
Join \(OP\), \(OQ\), and \(OR\).
Proof
Since \(PQ\) and \(PR\) are tangents:
\[ OQ \perp PQ \quad \text{and} \quad OR \perp PR \]
In right triangles \(\triangle OQP\) and \(\triangle ORP\):
- \(OQ = OR\) (radii of the same circle)
- \(OP = OP\) (common hypotenuse)
- \(\angle OQP = \angle ORP = 90^\circ\)
Therefore, \[ \triangle OQP \cong \triangle ORP \quad \text{(RHS)} \]
By CPCT, \[ PQ = PR \]
Additional Result
Also, \[ \angle OPQ = \angle OPR \] Hence, line \(OP\) bisects the angle between the two tangents.
📐 Derivation
Conceptual Insight
This theorem is based on RHS congruency of right triangles formed by radii and tangents. The perpendicular nature of radius to tangent ensures right angles, making congruency applicable.
✏️ Example
From a point \(P\), tangents \(PA\) and \(PB\) are drawn to a circle.
If \(PA = 6\text{ cm}\), find \(PB\).
Step by step Solution
- \[PA = PB \quad (\text{Tangents from same external point})\]
- \[PB = 6 \text{ cm}\]
⚡ Exam Tip
Exam Tips
⚠️ Warning
Common Mistakes
📋 Case Study
CBSE Case Study (HOTS)
A person standing outside a circular park walks equal distances along two paths that just touch the park boundary. Explain why these distances must be equal using circle geometry.
🌟 Importance
Importance for Board Exams
✏️ Example
Prove that in two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.
Chord AB of larger circle touches smaller circle at P
Given
Two concentric circles with centre \(O\).
\(AB\) is a chord of the larger circle.
\(AB\) touches the smaller circle at point \(P\).
\(AB\) is a chord of the larger circle.
\(AB\) touches the smaller circle at point \(P\).
To Prove
\[AP=AB\]
Construction
Join \(OA\), \(OB\), and \(OP\).
A tangent to a circle is perpendicular to the radius at the point of contact.
Proof
Since chord \(AB\) touches the smaller circle at \(P\), it acts as a tangent to the smaller circle.
Therefore, \[ OP \perp AB \]
In triangles \(\triangle OAP\) and \(\triangle OBP\):
- \(OA = OB\) (radii of larger circle)
- \(OP = OP\) (common side)
- \(\angle OPA = \angle OPB = 90^\circ\)
Hence, \[ \triangle OAP \cong \triangle OBP \quad (\text{RHS}) \]
By CPCT, \[ AP = PB \]
Conclusion
The chord of the larger circle is bisected at the point where it touches the smaller circle.
⚡ Exam Tip
Exam Tips
⚠️ Warning
Common Mistakes
📋 Case Study
Common Mistakes
- Missing that chord acts as tangent to smaller circle.
- Not proving perpendicular condition.
- Incorrect triangle pairing.
🌟 Importance
Why This Question is Important
✏️ Example
Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that \(\angle PTQ = 2 \angle OPQ.\)
Different positions of a point relative to a circle
Given
A circle with centre \(O\).
\(T\) is an external point.
\(TP\) and \(TQ\) are tangents at \(P\) and \(Q\).
TO Prove
\[ \angle PTQ = 2\angle OPQ \]
- Tangents from an external point are equal
- Radius is perpendicular to tangent
- Angle sum property of triangle
Proof
Since tangents from external point are equal: \[ TP = TQ \] Therefore, \(\triangle TPQ\) is isosceles.
Hence, \[ \angle TPQ = \angle TQP \]
Let \(\angle PTQ = \theta\)
Using angle sum property: \[ \theta + 2\angle TPQ = 180^\circ \]
\[ \Rightarrow \angle TPQ = 90^\circ - \frac{\theta}{2} \]
Now, since radius is perpendicular to tangent: \[ \angle OPQ + \angle TPQ = 90^\circ \]
Substituting: \[ \angle OPQ = 90^\circ - \left(90^\circ - \frac{\theta}{2}\right) \] \[ \Rightarrow \angle OPQ = \frac{\theta}{2} \]
Hence, \[ \angle PTQ = 2\angle OPQ \]
Alternative Insight
The angle between two tangents from an external point is equal to supplement of the angle subtended by the chord at the centre. This provides a shortcut method in objective questions.
⚡ Exam Tip
Exam Tips
⚠️ Warning
Common Mistakes
📋 Case Study
CBSE Case Study (HOTS)
Two roads touch a circular park from a common point. Analyze how the angle between roads relates to angles formed inside the park using circle geometry.
🌟 Importance
Importance for Board Exams
✏️ Example
PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T. Find the length TP.
Tangents at P and Q intersect at T
- Perpendicular from centre to chord bisects the chord
- Radius ⟂ tangent
- Similarity of triangles (AA)
Step by step Solution
- Find Distance from Centre to Chord
- Let \(R\) be midpoint of chord \(PQ\).
- \[PR = \frac{PQ}{2} = \frac{8}{2} = 4 \text{ cm}\]
- In right triangle \(\triangle OPR\):
- \[ \begin{aligned} OR^2 &= OP^2 - PR^2\\ &= 5^2 - 4^2 \\&= 25 - 16 \\&= 9\\ OR &= 3 \text{ cm} \end{aligned} \]
- Use Triangle Similarity
In triangles \(\triangle TPR\) and \(\triangle PRO\):
\(\angle TPR = 90^\circ\) (tangent ⟂ radius)
\(\angle PRO = 90^\circ\)
Common angle at \(R\)
Therefore, \[ \triangle TPR \sim \triangle PRO \quad (\text{AA}) \] - \[\frac{TP}{PO} = \frac{PR}{OR}\]
- \[ \begin{aligned} TP &= \frac{PR \cdot PO}{OR}\\ &= \frac{4 \times 5}{3}\\ &= \frac{20}{3} \text{ cm} \end{aligned} \]
Final Answer
\[ TP = \frac{20}{3} \text{ cm} \]
Smart Insight (Shortcut)
This problem can also be solved using Power of a Point, where: \[ TP^2 = \text{(distance from centre)}^2 - r^2 \] but similarity gives a cleaner CBSE-friendly method.
⚡ Exam Tip
Exam Tips
⚠️ Warning
Common Mistakes
🌟 Importance
Why This Question is Important
💡 Concept
Perpendicular from Centre to Chord
The perpendicular drawn from the centre of a circle to a chord bisects the chord.
\[ OP \perp AB \Rightarrow AP = PB \]
Derivation Insight
- Join \(OA\) and \(OB\)
- Use RHS congruency in triangles
Reverse Theorem
If a line from the centre bisects a chord, then it is perpendicular to the chord.
Exam Tip
This theorem is frequently used in hidden form in proofs.
💡 Concept
Equal Chords and Equal Distances
Equal chords of a circle are equidistant from the centre.
\[ AB = CD \Rightarrow OP = OQ \]
Converse
Chords equidistant from the centre are equal.
Usage
- Comparing lengths indirectly
- Used in symmetry-based proofs
💡 Concept
Tangent-Secant Relation (Advanced Insight)
If a tangent and a secant are drawn from an external point:
\[ (\text{Tangent})^2 = (\text{External part}) \times (\text{Whole secant}) \]
Why Important
- Foundation of Power of a Point
- Appears in Olympiad and advanced CBSE questions
💡 Concept
Power of a Point (Master Concept)
The power of a point defines a constant product relation for lines drawn from a point to a circle.
\[ PA \cdot PB = PC \cdot PD = (\text{Tangent})^2 \]
Cases
- Two secants
- One tangent + one secant
- Two tangents
Exam Insight
This concept unifies multiple theorems into one framework.
💡 Concept
Angle Properties in Circles
- Angle between tangent and radius = \(90^\circ\)
- Angle between tangent and chord equals angle in alternate segment
- Angle between two tangents: \[ = 180^\circ - \text{angle at centre} \]
hortcut Insight
Use central angle relations to avoid long proofs.
💡 Concept
High-Utility Results
- Radius ⟂ tangent at point of contact
- Tangents from external point are equal
- Line joining centre to midpoint of chord is perpendicular
- Chord nearer to centre is longer
💡 Concept
Problem Solving Roadmap
- Identify tangent / chord / secant
- Mark right angles (very important)
- Check symmetry or equal lengths
- Apply RHS or similarity
- Use power of a point if lengths involved
💡 Concept
CBSE Question Pattern Analysis
⚠️ Warning
Common Errors to Avoid
🌟 Importance
Ultra-Fast Revision Sheet
NCERT · Class X · Chapter 10
Circles — AI Learning Engine
Theorems · Proofs · Step-by-Step Solver · Interactive Explorations
Chapter 10 explores the elegant relationship between circles and lines — where tangents, secants, and radii meet to reveal beautiful geometric truths used in real-world applications from architecture to engineering.
Theorems
2 Core
Key Formulas
8 Types
Practice Qs
24 Problems
Modules
5 Interactive
🗺 Chapter Roadmap
From basic definitions to advanced theorem applications
1. Recalling Circle Vocabulary
Centre, radius, diameter, chord, arc, sector, tangent, secant — precise language for precise reasoning.
2. Lines and Circles — Three Cases
A line can be non-intersecting, a secant (cuts at 2 points), or a tangent (touches at exactly 1 point — the point of contact).
3. Theorem 10.1 — Perpendicularity
The radius to a point of tangency is perpendicular to the tangent. This is the master key to nearly every Circle problem.
4. Tangents from an External Point
Exactly two tangents can be drawn from any external point, and they are always equal in length (Theorem 10.2).
5. Angle Relationships & Applications
∠OPA = ∠OPB = 90°, ∠AOB + ∠APB = 180°, and the beautiful consequences for triangles, quadrilaterals, and more.
🔑 Why This Chapter Matters
🏗
Engineering
Gear teeth, cam profiles, and pipe intersections use tangent geometry constantly.
🎨
Art & Architecture
Islamic geometric patterns and arch design rely on tangent circle constructions.
📡
GPS & Navigation
Tangent lines model the horizon — the limit of visible Earth from a satellite.
📐 Core Concepts
Click any card to expand the full explanation, proof, and worked insight.
Concept 01
Tangent to a Circle
A tangent is a line that touches a circle at exactly one point — the point of tangency or point of contact.
A tangent to a circle is a line that intersects the circle in exactly one point.
Unlike a secant (which intersects at two points), the tangent "grazes" the circle. The number of tangents depends on where the point lies:
🔵 Point inside circle: No tangent possible — any line through it will be a secant.
🟡 Point on the circle: Exactly one tangent — perpendicular to the radius at that point.
🟢 Point outside the circle: Exactly two tangents — both equal in length.
This three-case classification completely characterises tangency for any point in the plane.
Concept 02 · Theorem 10.1
Tangent ⊥ Radius at Point of Contact
The radius drawn to the point of tangency is always perpendicular to the tangent line. This is the most-used fact in this chapter.
Theorem 10.1: The tangent at any point of a circle is perpendicular to the radius through the point of contact.
Proof (by contradiction):
Given: Circle with centre O, tangent PQ at point P.
To Prove: OP ⊥ PQ.
Assume OP is NOT perpendicular to PQ. Then there exists a point R on PQ such that OR ⊥ PQ, and OR < OP.
But R lies on PQ (tangent), which means R is outside the circle. So OR must be greater than the radius: OR > OP. Contradiction!
Hence our assumption is false. Therefore OP ⊥ PQ. ∎
Consequence: In right △OAP, we can always use Pythagoras' theorem — OP² = OA² + AP².
Concept 03 · Theorem 10.2
Equal Tangents from External Point
From any external point, exactly two tangents can be drawn to a circle, and these two tangent segments are equal in length.
Theorem 10.2: The lengths of the two tangents drawn from an external point to a circle are equal.
Proof (using RHS congruence):
Given: External point P; tangents PA and PB with A, B on the circle; O is the centre.
In △OAP and △OBP:
OA = OB (radii of same circle)
OP = OP (common hypotenuse)
∠OAP = ∠OBP = 90° (radius ⊥ tangent, Theorem 10.1)
OA = OB (radii of same circle)
OP = OP (common hypotenuse)
∠OAP = ∠OBP = 90° (radius ⊥ tangent, Theorem 10.1)
By RHS congruence: △OAP ≅ △OBP
Therefore PA = PB. ∎
Bonus: ∠OPA = ∠OPB (CPCT), so OP bisects angle APB. Also ∠AOB + ∠APB = 180°.
Concept 04
Length of Tangent Formula
Given distance from external point P to centre O, and radius r, the tangent length can be calculated using Pythagoras.
Tangent Length = √(OP² − r²)
Derivation: Since ∠OAP = 90° (Theorem 10.1), right △OAP gives:
OP² = OA² + AP²
OP² = r² + PA²
PA² = OP² − r²
PA = √(OP² − r²)
OP² = r² + PA²
PA² = OP² − r²
PA = √(OP² − r²)
Valid only when P is external, i.e., OP > r. Also: ∠APO = arctan(r / PA).
Concept 05
Angle Between Tangent Pair
The angle at the external point between the two tangents and the angle at the centre between the two radii to tangency points are supplementary.
∠APB + ∠AOB = 180°
In quadrilateral OAPB:
∠OAP = 90°, ∠OBP = 90° (tangent ⊥ radius)
Sum of angles = 360°
90° + 90° + ∠AOB + ∠APB = 360°
∠AOB + ∠APB = 180°
∠OAP = 90°, ∠OBP = 90° (tangent ⊥ radius)
Sum of angles = 360°
90° + 90° + ∠AOB + ∠APB = 360°
∠AOB + ∠APB = 180°
Also: OP bisects ∠APB and ∠AOB. Each of ∠OPA = ∠OPB = ½ ∠APB.
Concept 06
Tangent-Chord Angle
The angle between a tangent and a chord drawn from the point of tangency equals the inscribed angle in the alternate segment.
Tangent-Chord angle = Inscribed angle in alternate segment (Alternate Segment Theorem)
If tangent PT at P meets chord PQ, then ∠TPQ = ∠PRQ where R is any point on the major arc PQ.
Key use: This elegantly links tangent angles to inscribed angles, reducing complex circle problems to simple angle chasing.
This theorem (also called the Tangent-Chord Angle Theorem) extends Chapter 10 beautifully into Chapter 9 concepts.
📋 Formula Reference
Every formula you need — with context, variable meanings, and usage notes.
Formula 01
Length of Tangent
PT = √(PO² − r²)
P = external point, O = centre, r = radius, PT = tangent length. Requires PO > r (P outside circle).
Formula 02
Pythagoras in Tangent Triangle
PO² = PT² + r²
Rearrangement of Formula 01. Useful when finding PO or r. Right angle is at T (point of tangency).
Formula 03
Equal Tangents
PA = PB
P = external point, A and B = points of tangency. The two tangent segments from any external point are always equal.
Formula 04
Supplementary Angle Pair
∠AOB + ∠APB = 180°
O = centre, P = external point, A and B = tangency points. Central angle and external angle are supplementary.
Formula 05
Bisection by OP
∠OPA = ∠OPB = ½ ∠APB
The line from centre to external point bisects the angle between the tangents AND the angle at the centre.
Formula 06
Tangent Perpendicularity
∠OAP = ∠OBP = 90°
Both radii to points of tangency are perpendicular to respective tangents. Foundation of all tangent calculations.
Formula 07
Tangent Angle at Centre
∠AOB = 180° − ∠APB
Direct rearrangement of Formula 04. If ∠APB = 60°, then ∠AOB = 120°. Useful for finding central angle.
Formula 08
Tangent from Point on Circle
PT = 0 (one tangent only)
When P lies ON the circle, OP = r, so PT = √(r² − r²) = 0. Only one tangent exists at that point.
🗺 Quick Formula Map
Finding Tangent Length
Use PT = √(PO²−r²). Need: distance from point to centre, and radius.
Finding an Angle
Use ∠AOB + ∠APB = 180°. Draw the quadrilateral OAPB first.
Proving Equality
Use PA = PB (Theorem 10.2). Prove via RHS congruence in △OAP and △OBP.
Triangle Problems
Circumscribed triangle: use equal tangent property at each vertex to set up equations.
💡 Tips, Tricks & Common Mistakes
Smart shortcuts and the errors that cost marks in exams.
✨ Tips & Tricks
Draw First, Think Later. Always construct the complete figure — mark radii, right angles, and label all given points before writing a single equation.
The Right-Angle Radar. Every time you see "tangent" in a problem, immediately mark ∠OAP = 90°. This reflex alone solves 80% of problems.
Two Tangents = Equal Pair. From any external point P, PA = PB always. Use this to replace unknowns with known values instantly.
Quadrilateral OAPB. The four-sided figure formed by two radii, two tangents, and the line OP has angle sum 360°. Always use this for angle problems.
Circumscribed Triangle Shortcut. If a triangle ABC is circumscribed about a circle, then: AB + CD = AC + BD (equal tangents at each vertex). Saves enormous time!
OP is the Key Line. The line from centre O to external point P bisects ∠APB, bisects ∠AOB, and is the axis of symmetry of the entire configuration.
Pythagoras is Always Available. In right △OAP, you have three sides (OP, OA = r, PA = tangent). Any one unknown can be found from the other two.
⚠ Common Mistakes to Avoid
Assuming tangent = chord. A tangent touches at ONE point only. A chord connects two points. Never confuse their lengths or angle properties.
Forgetting the right angle. Many students forget ∠OAP = 90° mid-solution and use the wrong formula. Mark it explicitly at the start.
PA = PB means lengths, not vectors. PA and PB are equal in length, but they point in different directions. ∠PAB ≠ 0°.
∠AOB + ∠APB = 360°. Wrong! The correct relation is ∠AOB + ∠APB = 180°. The 360° is for the full quadrilateral OAPB.
Applying tangent formula to internal points. PT = √(PO² − r²) is only valid when P is outside the circle. When PO < r, the expression is imaginary — no tangent exists.
Saying "two tangents from a point on the circle". From a point ON the circle, there is exactly one tangent (perpendicular to the radius at that point).
Skipping RHS congruence justification. In proofs, you must cite the congruence criterion (RHS here) explicitly. Just writing "triangles are congruent" loses marks.
🧮 Step-by-Step AI Solver
Describe any Circle problem from Chapter 10. The engine will identify the concept, cite the theorem, and walk through every step.
📝 Problem Input
Quick examples:
✏ Practice Questions
24 original concept-building problems — organised by topic with full step-by-step solutions. Click any question to reveal the solution.
⚡ Interactive Learning Modules
Five hands-on tools for concept building through exploration.
🔵 Tangent Explorer
Drag the external point or adjust the radius. Watch how tangents, angles, and lengths update in real time.
Radius (r): 60 px
External Distance: —
Radius (r): 60 px
Distance OP: —
Tangent length PA: —
∠OAP: 90°
∠AOB: —
∠APB: —
Click on canvas to move external point P
📜 Visual Proof — Theorem 10.2
Click each step to highlight it on the diagram. Watch the proof come alive.
1
O is the centre. P is an external point. PA and PB are tangents, touching the circle at A and B respectively.
2
Draw radii OA and OB. By Theorem 10.1, ∠OAP = ∠OBP = 90° (radius ⊥ tangent at point of contact).
3
Draw line OP. Now consider △OAP and △OBP. OA = OB (radii), OP = OP (common side).
4
Both triangles have a right angle: ∠OAP = ∠OBP = 90°. By RHS congruence criterion: △OAP ≅ △OBP.
5
By CPCT: PA = PB. The two tangent segments from external point P are equal in length. ∎
🧠 Quick Quiz — 10 Questions
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Class 10 Maths Ch 10 Notes: Circles (Easy & Quick Revision)
Class 10 Maths Ch 10 Notes: Circles (Easy & Quick Revision) — Complete Notes & Solutions · academia-aeternum.com
The chapter “Circles” in NCERT Mathematics Class X introduces learners to one of the most elegant and logically rich areas of geometry. Building upon earlier concepts of lines, angles, and triangles, this chapter focuses on understanding the geometric behaviour of circles through precise definitions, carefully reasoned theorems, and visual interpretation. Students explore fundamental ideas such as tangents, secants, chords, points of contact, and the unique relationships that exist between a…
🎓 Class 10
📐 Mathematics
📖 NCERT
✅ Free Access
🏆 CBSE · JEE
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