Key Theorems Must Learn
The tangent at any point of a circle is perpendicular to the radius through the point of contact.
OP ⟂ Tangent at P
Given: A circle with centre O. PT is a tangent at point P on the circle.
To Prove: OP ⟂ PT
Proof (by contradiction): Assume OP is not perpendicular to PT. Draw OQ ⟂ PT. Q lies outside the circle (since tangent touches at only one point). In right Î"OQP, OQ < OP (hypotenuse is longest). But OP = radius, so OQ < radius, meaning Q is inside the circle. This contradicts that PT is a tangent. ∴ OP ⟂ PT.
Conclusion: ∠OPT = 90° – this fact is used in almost every problem.
The lengths of tangents drawn from an external point to a circle are equal.
PA = PB
Given: P is an external point. PA and PB are tangents to the circle from P.
To Prove: PA = PB
Proof: Join OA, OB, and OP.
In Î"OAP and Î"OBP:
• OA = OB (radii of same circle)
• OP = OP (common side)
• ∠OAP = ∠OBP = 90° (Th 10.1 – radius ⟂ tangent)
∴ Î"OAP ≅ Î"OBP (by RHS criterion)
∴ PA = PB (CPCT) âœ"
Also, ∠OPA = ∠OPB (tangents are equally inclined to line joining centre to external point).
Exercise 10.1 4 Questions – Basic Concepts
How many tangents can a circle have?
A circle has infinitely many tangents.
Explanation: A tangent touches the circle at exactly one point. Since there are infinitely many points on the circumference of a circle, and at each point you can draw exactly one tangent (perpendicular to the radius at that point), the number of tangents to a circle is infinite.
Fill in the blanks:
(i) A tangent to a circle intersects it in one point(s).
(ii) A line intersecting a circle in two points is called a secant.
(iii) A circle can have two parallel tangents at the most.
(iv) The common point of a tangent to a circle and the circle is called the point of contact.
At most 2 parallel tangents
Secant intersects at 2 points
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Find the length PQ.
Right triangle OPQ
Given: Radius OP = 5 cm, OQ = 12 cm. PQ is tangent at P.
Step 1: By Theorem 10.1, OP ⟂ PQ. So ∠OPQ = 90°.
Step 2: In right triangle OPQ, by Pythagoras theorem:
OQ² = OP² + PQ²
Step 3: 12² = 5² + PQ²
144 = 25 + PQ²
PQ² = 144 âˆ' 25 = 119
Step 4: PQ = √119 ≈ 10.9 cm
Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.
l₠⟂ l ⟂ lâ‚‚ – All three are parallel
Steps:
1. Draw a circle with centre O.
2. Draw the given line l.
3. Draw a line lâ‚ parallel to l that touches the circle at exactly one point – this is the tangent.
4. Draw another line lâ‚‚ parallel to l that cuts the circle at two points – this is the secant.
Note: A tangent line can be parallel to a secant line. Both are parallel to the given direction.
Exercise 10.2 13 Questions – Tangents from an External Point
Theorem 10.1: Tangent ⟂ Radius at point of contact (∠OPT = ∠OQT = 90°).
Theorem 10.2: Tangents from an external point are equal (TP = TQ).
Pythagoras Theorem: Used when we know two sides of a right triangle.
From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. Find the radius of the circle.
Right triangle OPQ
Given: Tangent length PQ = 24 cm, OQ = 25 cm (distance from centre to Q).
Step 1: Since PQ is tangent at P, OP ⟂ PQ (Theorem 10.1). So ∠OPQ = 90°.
Step 2: In right Î"OPQ:
OQ² = OP² + PQ² (Pythagoras)
25² = r² + 24²
625 = r² + 576
r² = 625 âˆ' 576 = 49
Answer: r = 7 cm
Check: 7² + 24² = 49 + 576 = 625 = 25² âœ"
In the given figure, TP and TQ are two tangents to a circle with centre O such that ∠POQ = 110°. Find ∠PTQ.
∠POQ = 110°
Given: TP and TQ are tangents from T. ∠POQ = 110°.
Step 1: By Theorem 10.1, ∠OPT = 90° and ∠OQT = 90°.
Step 2: In quadrilateral OPTQ, sum of all angles = 360°.
∠OPT + ∠OQT + ∠POQ + ∠PTQ = 360°
90° + 90° + 110° + ∠PTQ = 360°
290° + ∠PTQ = 360°
Answer: ∠PTQ = 70°
Key insight: ∠PTQ and ∠POQ are supplementary (sum to 180°).
If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80°, then find ∠POA.
∠APB = 80°
Given: PA & PB are tangents, ∠APB = 80°. Need ∠POA.
Step 1: By Theorem 10.2, PA = PB. By Theorem 10.1, ∠PAO = 90°.
Step 2: OP bisects ∠APB (from congruence of Î"OAP & Î"OBP).
So ∠APO = ∠BPO = 80°/2 = 40°.
Step 3: In right triangle OAP:
∠POA + ∠APO = 90° (complementary in right triangle)
∠POA + 40° = 90°
Answer: ∠POA = 50°
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Tangents at ends of diameter AB
Given: AB is a diameter of the circle with centre O. Tangent l at A, tangent m at B.
To Prove: l ∥ m (tangents are parallel).
Proof:
• By Theorem 10.1: Radius OA ⟂ Tangent l at A â‡' ∠OAl = 90°
• By Theorem 10.1: Radius OB ⟂ Tangent m at B â‡' ∠OBm = 90°
• Since AB is a diameter, OA and OB are parts of the same straight line.
• So both tangent l and tangent m are perpendicular to the same line AB.
• Lines perpendicular to the same line are parallel to each other.
∴ l ∥ m âœ"
Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Given: A circle with centre O. PT is a tangent at point P. PQ is a line drawn perpendicular to PT at P.
To Prove: PQ passes through the centre O.
Proof:
• By Theorem 10.1, the radius OP ⟂ PT at the point of contact P.
• So OP is a line through P that is perpendicular to PT.
• There can be only one line through a given point perpendicular to a given line.
• Since PQ is also drawn perpendicular to PT at P, PQ must be the same line as OP.
• Therefore, PQ passes through the centre O.
∴ The perpendicular at the point of contact to the tangent passes through the centre. âœ"
Corollary: The shortest distance from the centre to the tangent is the radius drawn to the point of contact.
The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.
Given: OA = 5 cm (distance from centre to point A), tangent AP = 4 cm.
Step 1: Let the point of contact be P. By Theorem 10.1, OP ⟂ AP.
Step 2: In right triangle OAP:
OA² = OP² + AP²
5² = r² + 4²
25 = r² + 16
r² = 9
Answer: r = 3 cm
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.
Chord AB touches inner circle at M
Given: Two concentric circles (same centre O). Larger radius = 5 cm, smaller radius = 3 cm.
Step 1: Let AB be the chord of the larger circle. It touches the smaller circle at M, so AB is tangent to the smaller circle at M.
Step 2: By Theorem 10.1, OM ⟂ AB. So M is the midpoint of AB (perpendicular from centre to a chord bisects the chord).
Step 3: In right triangle OMA:
OA² = OM² + AM²
5² = 3² + AM²
25 = 9 + AM² â‡' AM² = 16 â‡' AM = 4 cm
Step 4: AB = 2 × AM = 8 cm
A quadrilateral ABCD is drawn to circumscribe a circle (see diagram). Prove that AB + CD = AD + BC.
Quadrilateral circumscribing a circle
Given: ABCD circumscribes a circle. The circle touches AB, BC, CD, DA at P, Q, R, S respectively.
Proof: Using Theorem 10.2 (tangents from an external point are equal):
• From point A: AP = AS (tangents to the circle)
• From point B: BP = BQ
• From point C: CQ = CR
• From point D: DR = DS
Now:
AB + CD = (AP + BP) + (CR + DR)
= (AS + BQ) + (CQ + DS) [substituting equal lengths]
= (AS + DS) + (BQ + CQ)
= AD + BC âœ"
In the figure, XY and X′Y′ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X′Y′ at B. Prove that ∠AOB = 90°.
Given: XY ∥ X′Y′. Another tangent AB touches at C, meeting XY at A and X′Y′ at B.
Proof: Join OC, OA, OB, OP, OQ.
• In Î"OPA and Î"OCA: OP = OC (radii), OA = OA, AP = AC (tangents from A). ∴ Î"OPA ≅ Î"OCA (SSS). So ∠1 = ∠2 (OA bisects ∠POC).
• Similarly, Î"OCB ≅ Î"OQB (SSS). So ∠3 = ∠4 (OB bisects ∠COQ).
• Now, ∠POQ = 180° (P, O, Q are collinear since XY ∥ X′Y′ and P, Q are points of contact).
• ∠POC + ∠COQ = 180°
â‡' 2∠2 + 2∠3 = 180° â‡' ∠2 + ∠3 = 90°
∴ ∠AOB = 90° âœ"
Prove that 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.
Prove: ∠APB + ∠AOB = 180°
Given: PA and PB are tangents from external point P. A and B are points of contact.
To Prove: ∠APB + ∠AOB = 180° (supplementary)
Proof using Quadrilateral OAPB:
• ∠OAP = 90° (radius OA ⟂ tangent PA – Theorem 10.1)
• ∠OBP = 90° (radius OB ⟂ tangent PB – Theorem 10.1)
• Sum of all angles in quadrilateral OAPB:
∠OAP + ∠OBP + ∠APB + ∠AOB = 360°
90° + 90° + ∠APB + ∠AOB = 360°
∠APB + ∠AOB = 180° âœ"
Note: This is a very important result – often tested in CBSE exams.
Prove that the parallelogram circumscribing a circle is a rhombus.
Given: ABCD is a parallelogram circumscribing a circle.
Proof:
• ABCD is a parallelogram â‡' AB = CD and AD = BC (opposite sides equal).
• From Q8 (quadrilateral circumscribing circle): AB + CD = AD + BC.
• Substitute AB = CD and AD = BC:
AB + AB = AD + AD
2AB = 2AD
â‡' AB = AD
• A parallelogram with adjacent sides equal is a rhombus.
∴ ABCD is a rhombus âœ"
A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC.
Given: Circle touches BC at D, CA at E, AB at F. BD = 8 cm, DC = 6 cm, radius = 4 cm.
Step 1: Let lengths be: BD = BF = 8, DC = CE = 6, AF = AE = x (tangents from A).
Step 2: Sides: BC = 14, AB = 8 + x, AC = 6 + x.
Step 3: Area of Î"ABC = Area of Î"OBC + Î"OCA + Î"OAB (all have height = radius = 4).
ar(ABC) = ½(BC)(r) + ½(CA)(r) + ½(AB)(r) = ½r(BC + CA + AB)
ar(ABC) = ½(4)(14 + 6 + x + 8 + x) = 2(28 + 2x) = 56 + 4x
Step 4: Using semi-perimeter s = (14 + 14 + 2x)/2 = 14 + x and Heron's formula, or equating with s × r: ar = s × r = (14 + x) × 4 = 56 + 4x âœ"
Also using ar = ½ × BC × height from A â†' gives x = 7. So AB = 15 cm, AC = 13 cm
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Given: Quadrilateral ABCD circumscribes a circle with centre O. The sides AB, BC, CD, DA touch the circle at P, Q, R, S respectively.
To Prove: ∠AOB + ∠COD = 180° and ∠BOC + ∠AOD = 180°
Proof:
• In Î"AOP and Î"AOS: AP = AS (tangents from A), OP = OS (radii), OA = OA. So Î"AOP ≅ Î"AOS (SSS).
∴ ∠AOP = ∠AOS. Let each = ∠1.
• Similarly, from congruence at other vertices:
∠BOP = ∠BOQ = ∠2, ∠COQ = ∠COR = ∠3, ∠DOR = ∠DOS = ∠4.
• Sum of angles around O: 2(∠1 + ∠2 + ∠3 + ∠4) = 360°
â‡' ∠1 + ∠2 + ∠3 + ∠4 = 180°
• ∠AOB = ∠1 + ∠2 and ∠COD = ∠3 + ∠4
∴ ∠AOB + ∠COD = ∠1 + ∠2 + ∠3 + ∠4 = 180° âœ"
Similarly, ∠BOC + ∠AOD = 180° âœ"
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