Chapter 10 — Circles

Theory Used

A fundamental property of tangents to a circle states that:

The tangent drawn from an external point to a circle is perpendicular to the radius at the point of contact.

Therefore, triangle formed is a right-angled triangle and we apply:

\[ \text{Pythagoras Theorem: } (\text{Hypotenuse})^2 = (\text{Base})^2 + (\text{Perpendicular})^2 \]

Solution Roadmap

• Identify right triangle formed by radius, tangent, and line joining centre to external point
• Use Pythagoras theorem
• Substitute given values
• Compute radius

Solution

Let O be the centre, P the point of contact and Q the external point.

Given:
Length of tangent \( PQ = 24 \, cm \)
Distance \( OQ = 25 \, cm \)

Since radius is perpendicular to tangent:

\[ OP \perp PQ \]

Therefore, triangle \( OPQ \) is right-angled at P.

Applying Pythagoras theorem:

Substitute values:

Rearranging:

Hence, radius of the circle is:

7 cm

Correct Option: (A)

Significance for Exams

• This is a standard CBSE board MCQ based on tangent-radius property
• Frequently asked in competitive exams like NDA, SSC, Railways
• Tests conceptual clarity of right-angle formation in circles
• Also builds foundation for advanced geometry problems involving power of a point

Theory Used

The angle between two tangents drawn from an external point is given by:

This follows because:

• Radius is perpendicular to tangent at point of contact
• A quadrilateral is formed with two right angles
• Sum of angles in quadrilateral = \(360^\circ\)

Solution Roadmap

• Identify quadrilateral OPQT
• Use right angle property at points of contact
• Apply angle sum of quadrilateral
• Solve for required angle

Solution

Given:
\( \angle POQ = 110^\circ \)

Since radius is perpendicular to tangent:

Consider quadrilateral \( OPQT \)

Sum of interior angles:

Substitute values:

Solve:

Final Answer: 70°

Correct Option: (B)

Significance for Exams

• Very common CBSE MCQ and 2-mark question pattern
• Direct application of tangent-angle theorem
• Frequently asked in NDA, SSC, Banking exams
• Shortcut formula \(180^\circ - \text{central angle}\) saves time in MCQs

Q3. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of \(80^\circ\), then \(\angle POA\) is equal to
(A) 50°
(B) 60°
(C) 70°
(D) 80°

Theory Used

• Length of tangents from an external point are equal: \(PA = PB\)
• Radius is perpendicular to tangent at point of contact
• Triangles formed are congruent (RHS)
• Line joining centre to external point bisects angle between tangents

Solution Roadmap

• Prove triangles \( \triangle PAO \cong \triangle PBO \)
• Use congruency to show angle bisector property
• Apply angle sum property in triangle
• Find required angle

Solution

Given:
\( \angle APB = 80^\circ \)

Since tangents from an external point are equal:

Also:

Therefore, in triangles \( \triangle PAO \) and \( \triangle PBO \):

Hence,

By CPCT:

Since total angle at P is \(80^\circ\):

Now in triangle \( \triangle PAO \):

Substitute values:

Solving:

Final Answer: 50°

Correct Option: (A)

Significance for Exams

• Tests understanding of tangent properties + triangle congruency
• Frequently asked in CBSE 2–3 mark questions
• Important for NDA, SSC geometry sections
• Angle bisection by line joining centre is a key shortcut concept

Q4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Theory Used

• The tangent at any point of a circle is perpendicular to the radius through the point of contact
• If two lines are perpendicular to the same line, then they are parallel to each other

Proof Roadmap

• Identify diameter and radii
• Apply perpendicular property of tangent
• Use parallel line criterion

Construction

Draw a circle with centre O. Let AB be a diameter of the circle. Draw tangents at points A and B.

Proof

Let AB be the diameter of the circle and O be its centre.

Draw tangents at points A and B.

Since radius is perpendicular to tangent at the point of contact:

But OA and OB lie on the same straight line AB (since AB is diameter).

Therefore, both tangents are perpendicular to the same line AB.

Using the theorem:

Hence,

Hence Proved

Significance for Exams

• Classic CBSE proof-based question (2–3 marks)
• Tests understanding of perpendicularity and parallelism
• Very common in board exams and school-level competitions
• Concept used in coordinate geometry and higher constructions

Q5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Theory Used

• The tangent at any point of a circle is perpendicular to the radius at the point of contact
• Through a given point on a line, only one perpendicular can be drawn

Proof Roadmap

• Use tangent ⟂ radius theorem
• Apply uniqueness of perpendicular at a point
• Conclude both lines coincide

Given

A circle with centre O and a tangent AB touching the circle at point P.

To Prove

The perpendicular at point P to the tangent AB passes through the centre O.

Construction

Join OP.

Proof

By the property of tangents:

Thus, OP is perpendicular to AB at point P.

Also, from geometry:

Therefore, the perpendicular drawn at point P to AB must be the line OP itself.

Since OP passes through O,

The perpendicular at P to the tangent passes through the centre O.

Hence Proved

Significance for Exams

• Fundamental theorem frequently used in proofs and constructions
• Appears in CBSE 2-mark and 3-mark questions
• Base concept for tangent-based locus and coordinate geometry problems
• Helps avoid common misconception: tangent is not arbitrary, it is uniquely defined

Theory Used

The tangent at any point of a circle is perpendicular to the radius at the point of contact.

Hence, triangle formed is right-angled and:

Solution Roadmap

• Identify right triangle \( \triangle AOB \)
• Apply Pythagoras theorem
• Substitute values
• Solve for radius

Solution

Let O be the centre and B be the point of contact.

Given:
\( AO = 5 \, cm \)
\( AB = 4 \, cm \)

Since radius is perpendicular to tangent:

Therefore, triangle \( \triangle AOB \) is right-angled at B.

Applying Pythagoras theorem:

Substitute values:

Rearranging:

Taking square root:

Radius of the circle = 3 cm

Significance for Exams

• Direct application of tangent-radius perpendicular property
• Very common CBSE 1–2 mark numerical question
• Frequently appears in MCQs of SSC, NDA, Railways
• Fast-solving question using right triangle recognition

Q7. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Theory Used

• A chord touching a circle is a tangent to that circle
• The radius drawn to the point of contact is perpendicular to the tangent
• The perpendicular from centre to a chord bisects the chord

Solution Roadmap

• Identify that chord is tangent to inner circle
• Use perpendicular property from centre
• Apply Pythagoras theorem in right triangle
• Double the half-chord to get full length

Solution

Let O be the common centre and PQ be the chord of the larger circle touching the smaller circle at R.

Given:
Radius of larger circle \( OP = 5 \, cm \)
Radius of smaller circle \( OR = 3 \, cm \)

Since chord PQ touches the inner circle, it is tangent at point R.

Therefore:

Also, perpendicular from centre to chord bisects the chord.

Let PR be half of the chord.

In right triangle \( \triangle OPR \):

Substitute values:

Rearranging:

Therefore, full chord:

Length of chord = 8 cm

Significance for Exams

• Classic application of tangent concept in concentric circles
• Frequently asked CBSE 3-mark problem
• Important for competitive exams (SSC, NDA)
• Tests multi-concept linking: tangent + chord + Pythagoras

Theory Used

• Tangents drawn from an external point to a circle are equal in length
• A circumscribed quadrilateral has all sides touching the circle

Proof Roadmap

• Mark points of contact on each side
• Apply equal tangent property at each vertex
• Express all sides in terms of tangent segments
• Add and rearrange to prove equality

Construction

Let the circle touch sides AB, BC, CD and DA at points P, Q, R and S respectively.

Proof

From the property of tangents:

Now, express sides:

Similarly:

Adding:

Rearranging:

Now:

Therefore:

Hence Proved

Significance for Exams

• Very important CBSE proof (3–4 marks)
• Frequently repeated in board exams
• Core concept in tangential quadrilaterals
• Useful in Olympiad and coordinate geometry applications

Theory Used

• Tangents from an external point are equal
• Line joining centre to external point bisects angle between tangents
• Interior angles on same side of transversal (parallel lines)
• Angle sum property of triangle

Solution Roadmap

• Use parallel line property for angles at A and B
• Apply tangent symmetry (angle bisector)
• Form equation using linear pair
• Use triangle angle sum to find required angle

Solution

Since XY ∥ X'Y' and AB is a transversal:

(Interior angles on same side of transversal)

From tangent properties:

(OA bisects angle between tangents at A)

(OB bisects angle between tangents at B)

Substitute in (1):

Now in triangle \( \triangle AOB \):

Using (2):

Hence Proved

Significance for Exams

• Advanced CBSE proof combining multiple theorems
• Tests deep understanding of tangents + parallel lines
• Frequently appears in 3–4 mark long-answer questions
• Important for Olympiad and geometry-heavy exams

Theory Used

• The tangent at any point is perpendicular to the radius through the point of contact
• Sum of interior angles of a quadrilateral is \(360^\circ\)

Proof Roadmap

• Form quadrilateral using centre, contact points and external point
• Use perpendicularity of radius and tangent
• Apply angle sum property

Given

AB and AC are tangents drawn from external point A to the circle with centre O.

To Prove

Construction

Join OB and OC.

Proof

Since radius is perpendicular to tangent:

Consider quadrilateral \( OBAC \)

Sum of interior angles:

Substitute values:

Rearranging:

Hence, the angle between the tangents is supplementary to the central angle.

Hence Proved

Significance for Exams

• One of the most important theorems of circles (frequent CBSE question)
• Forms base for many MCQs and long-answer proofs
• Widely used in competitive exams like NDA, SSC
• Shortcut result: \( \angle \text{between tangents} = 180^\circ - \text{central angle} \)

Theory Used

• Tangents drawn from an external point to a circle are equal
• In a parallelogram, opposite sides are equal

Proof Roadmap

• Mark points of contact
• Use equal tangent segments
• Express sides in segment form
• Compare adjacent sides

Given

ABCD is a parallelogram circumscribing a circle.

Construction

Let the circle touch sides AB, BC, CD and DA at points P, Q, R and S respectively.

Proof

From the property of tangents:

Express the sides:

Using (1) and (2):

Using (2) and (3):

Now from (1) and (3):

But more directly, from standard result (proved in previous question):

In a parallelogram:

Substitute:

Thus, adjacent sides are equal.

Also in parallelogram:

Therefore:

Hence, all sides are equal ⇒ ABCD is a rhombus

Hence Proved

Significance for Exams

• High-value CBSE proof (3–4 marks)
• Combines tangential quadrilateral + parallelogram properties
• Frequently asked in board exams
• Important for Olympiad geometry reasoning

Theory Used

• Radius to the point of contact is perpendicular to the tangent
• Sum of interior angles of a quadrilateral is \(360^\circ\)

Proof Roadmap

• Join radii to all points of contact
• Use perpendicular property (right angles)
• Form quadrilateral using centre and contact points
• Apply angle sum property

Given

ABCD is a quadrilateral circumscribing a circle with centre O.

Construction

Let the circle touch AB, BC, CD and DA at P, Q, R and S respectively. Join OP, OQ, OR and OS.

Proof

Since radius is perpendicular to tangent:

Therefore:

Consider quadrilateral \( PQRS \) with centre O joined.

Now focus on angles at centre subtended by opposite sides:

Side AB corresponds to angle \( \angle POS \)
Side CD corresponds to angle \( \angle QOR \)

In quadrilateral formed by joining O to P, Q, R, S:

Also, each adjacent pair forms right angles:

More directly, observe:

Angle between two radii corresponding to adjacent sides = 90° + 90° = 180°

Therefore:

Hence, opposite sides subtend supplementary angles at the centre.

Hence Proved

Significance for Exams

• Advanced geometry theorem (3–4 marks)
• Tests deep understanding of tangential quadrilateral
• Frequently asked in CBSE and Olympiad exams
• Builds foundation for cyclic vs tangential comparisons