Tangents and Intersecting Chords Chapter Notes | Mathematics Class 10 ICSE PDF Download

Introduction

• Imagine a circle and a straight line dancing together on a plane! 
• They can interact in three exciting ways:
  
• The line might shy away, not touching the circle at all (no contact).
• The line could boldly cut through the circle, meeting it at two points, known as a secant.
• The line might gently kiss the circle at just one point, called a tangent, creating a special point of contact.
• A secant is a line that intersects a circle at two points, like a sword slicing through.
• A tangent touches the circle at exactly one point, like a fleeting touch.
• The point where the tangent meets the circle is called the point of contact.

Theorem 10: Tangent and Radius are Perpendicular

Tangent-Radius Perpendicular

The tangent at any point on a circle is perpendicular to the radius passing through that point.

Stepwise Explanation:

• Consider a circle with center O. A tangent AB touches the circle at point P, and OP is the radius.
• Take another point Q on the tangent AB (not P) and draw line OQ.
• Since Q lies outside the circle, OP is shorter than OQ (the shortest distance from a point to a line is the perpendicular).
• Among all lines from O to AB, OP is the shortest, so OP is perpendicular to AB.

Key Points to Remember:

• No tangent can pass through a point inside the circle.
• Exactly one tangent can be drawn at a point on the circle’s circumference.
• Two tangents can be drawn from a point outside the circle.

Corollary: Properties of Two Tangents from an Exterior Point

When two tangents are drawn from an exterior point to a circle, they have special properties. 

Properties:

• The lengths of the two tangents are equal (PA = PB).
• The angles subtended by the tangents at the center are equal (∠AOP = ∠BOP).
• The tangents are equally inclined to the line joining the exterior point to the center (∠APO = ∠BPO).
• Consider a circle with center O. From an exterior point P, draw tangents PA and PB.
• In triangles AOP and BOP:
  • OA = OB (radii of the same circle).
  • ∠OAP = ∠OBP = 90° (radius is perpendicular to tangent).
  • OP = OP (common side).
  • Triangles AOP and BOP are congruent by RHS (Right angle-Hypotenuse-Side).
  • Thus, PA = PB, ∠AOP = ∠BOP, and ∠APO = ∠BPO (corresponding parts of congruent triangles).

Theorem 11: Point of Contact Lies on Line Through Centers

Case I: External Touching

• When two circles touch externally, their point of contact lies on the line joining their centers.
• Consider two circles with centers A and B, touching externally at point P.
• Draw a common tangent PQ through P and join AP and BP.
• ∠APQ = 90° and ∠BPQ = 90° (radius is perpendicular to tangent).
• ∠APQ + ∠BPQ = 180°, so ∠APB = 180°, meaning APB is a straight line.
• Thus, P lies on line AB.

Case II: Internal Touching

• When two circles touch internally, the point of contact lies on the line joining the centers, extended if necessary.
• Consider two circles with centers A and B, touching internally at point P.
• Since AP and BP are both perpendicular to PQ at P, they lie on the same line.
• Thus, P lies on the extended line AB.

Formulas for Distance Between Centers

For two circles with radii r1 and r2, and distance d between their centers:

• External touching: d = r₁ + r₂
• Internal touching: d = r₁ - r₂ (if r₁ > r₂) or d = r₂ - r₁ (if r₂ > r₁)

Additional Notes:

• If tangents AB and CD to a circle at points P and Q are parallel, then PQ is the diameter of the circle.
• Concentric circles share the same center.

Theorem 12: Product of Chord Segments

Case I: Internal Intersection

• When two chords intersect inside a circle, the product of the lengths of their segments is equal.
• Formula: PA × PB = PC × PD
• Consider chords AB and CD intersecting at point P inside the circle.
• Join AC and BD to form triangles APC and BPD.
• ∠A = ∠D and ∠C = ∠B (angles in the same segment).
• Triangles APC and BPD are similar by AA similarity.
• Thus, PA/PD = PC/PB, so PA × PB = PC × PD.

Case II: External Intersection

• When two chords, extended, intersect outside a circle, the product of the lengths of their segments is equal.
• Consider chords AB and CD, extended to intersect at point P outside the circle.
• Join AC and BD to form triangles PAC and PDB.
• ∠A = ∠PDB and ∠C = ∠PBD (exterior angle of a cyclic quadrilateral equals the interior opposite angle).
• Triangles PAC and PDB are similar by AA similarity.

Theorem 13: Angle Between Tangent and Chord

• The angle between a tangent and a chord at the point of contact equals the angle in the alternate segment.
• Formula: ∠BAQ = ∠ACB and ∠BAP = ∠ADB
• Stepwise Explanation:
  • Consider a circle with center O. Tangent PQ touches at A, and chord AB is drawn.
  • Draw diameter AOR and join RB.
  • In triangle ABR, ∠ABR = 90° (angle in a semicircle).
  • Thus, ∠ARB + ∠RAB = 90°.
  • Since ∠OAQ = 90° (radius perpendicular to tangent), ∠RAB + ∠BAQ = 90°.
  • Therefore, ∠ARB = ∠BAQ.
  • But ∠ARB = ∠ACB (angles in the same segment), so ∠BAQ = ∠ACB.
  • Since ∠BAP + ∠BAQ = 180° (straight line) and ∠ACB + ∠ADB = 180° (opposite angles of a cyclic quadrilateral), ∠BAP = ∠ADB.

Theorem 14: Tangent and Chord Intersection Externally

• When a chord and a tangent intersect externally, the product of the chord’s segment lengths equals the square of the tangent’s length from the point of contact to the intersection.
• Formula: PA × PB = PT²
• Stepwise Explanation:
  • Consider chord AB and tangent TP intersecting at P outside the circle.
  • Join TA and TB to form triangles PAT and PTB.
  • ∠PTB = ∠A (angle in the alternate segment).
  • ∠P = ∠P (common angle).
  • Triangles PAT and PTB are similar by AA similarity.
  • Thus, PA/PT = PT/PB, so PA × PB = PT².

Solved Examples

Example 1: Radius of Inscribed Circle In triangle PQR, PQ = 24 cm, QR = 7 cm, and ∠PQR = 90°. Find the radius of the inscribed circle.

Solution:

• Since ∠PQR = 90°, use Pythagoras: PR² = PQ² + QR² = 24² + 7² = 576 + 49 = 625, so PR = 25 cm.
• Let the incircle touch PQ at A, QR at B, and PR at C, with radius x cm.
• OAQB is a square, so AQ = BQ = x cm.
• PA = PQ - AQ = (24 - x) cm, RB = QR - BQ = (7 - x) cm.
• Tangents from an exterior point are equal: PC = PA = (24 - x) cm, RC = RB = (7 - x) cm.
• PR = PC + RC: 25 = (24 - x) + (7 - x) = 31 - 2x.
• Solve: 2x = 31 - 25 = 6, so x = 3.
• Radius of the inscribed circle = 3 cm.

Example 2: Equation for Radius of Third Circle Centers P and Q of circles with radii 9 cm and 2 cm are 17 cm apart. A third circle with center R and radius x cm touches both externally, with ∠PRQ = 90°. Find x.

Solution:

• For external touching, PR = 9 + x, QR = 2 + x, PQ = 17 cm.
• Since ∠PRQ = 90°, apply Pythagoras in triangle PRQ: PR² + QR² = PQ².
• (9 + x)² + (2 + x)² = 17².
• Expand: 81 + 18x + x² + 4 + 4x + x² = 289.
• Simplify: 2x² + 22x + 85 = 289, so 2x² + 22x - 204 = 0.
• Divide by 2: x² + 11x - 102 = 0.
• Factorize: (x + 17)(x - 6) = 0, so x = -17 or x = 6.
• Since radius cannot be negative, x = 6 cm.

Example 3: Length of Direct Common Tangent Two circles with radii 25 cm and 9 cm touch externally. Find the length of the direct common tangent.

Solution:

• Centers A and B, touching externally at O. MN is the direct common tangent.
• Draw BC ⊥ AM. BCMN is a rectangle, so BC = MN.
• AB = OA + OB = 25 + 9 = 34 cm.
• AC = AM - BN = 25 - 9 = 16 cm.
• In right triangle ABC, AB² = AC² + BC².
• 34² = 16² + BC², so 1156 = 256 + BC², BC² = 900, BC = 30 cm.
• Length of direct common tangent MN = BC = 30 cm.

Example 4: Length of Transverse Common Tangent Centers of two circles with radii 6 cm and 2 cm are 10 cm apart. Find the length of the transverse common tangent. 

Solution:

• Centers A and B, MN is the transverse common tangent.
• AB = 10 cm, AM = 6 cm, BN = 2 cm.
• Draw BC ⊥ AM extended. BCMN is a rectangle, so MN = BC, CM = BN = 2 cm.
• AC = AM + CM = 6 + 2 = 8 cm.
• In right triangle ABC, AB² = BC² + AC².
• 10² = BC² + 8², so 100 = BC² + 64, BC² = 36, BC = 6 cm.
• Length of transverse common tangent MN = BC = 6 cm.

Example 5: Angles in Triangle with Tangents In triangle PQR, PQ = QR, ∠RQP = 68°, PC and QC are tangents to a circle with center O. Find ∠QOP and ∠QCP. 

Solution:

• Since PQ = QR, triangle PQR is isosceles, so ∠QPR = ∠QRP.
• In triangle PQR, ∠QPR + ∠QRP + ∠RQP = 180°.
• Let ∠QPR = ∠QRP = x. Then x + x + 68° = 180°, so 2x = 112°, x = 56°.
• ∠QRP = 56°.
• Angle at center is twice the angle at circumference: ∠QOP = 2 × ∠QRP = 2 × 56° = 112°.
• In quadrilateral POQC, ∠QOP = 112°, ∠OPC = 90°, ∠OQC = 90° (radius ⊥ tangent).
• Sum of angles: ∠QOP + ∠OPC + ∠OQC + ∠QCP = 360°.
• 112° + 90° + 90° + ∠QCP = 360°, so ∠QCP = 68°.