Key Concepts
- 1Tangent perpendicularity
- 2Equal tangent lengths
- 3Tangents from external point
- 4∠PTQ + ∠POQ = ?
- 5Incircle tangent property
- 6Tangent from external point formula
- 7Tangent perpendicularity
- 8Tangents from external point
- 9Incircle tangent property
- 10What is a tangent to a circle?
Important Formulas & Facts
Tangent ⊥ radius at the point of contact.
Tangents from an external point to a circle are equal: PA = PB.
Exactly 2 tangents. From a point on the circle: 1. From inside: 0.
180° (since ∠OPT = ∠OQT = 90° in quad OPTQ)
For triangle with incircle: AP = AS, BP = BQ, CQ = CR (tangent pairs).
Length = √(d² - r²) where d = distance from point to centre.
Tangent ⊥ radius at the point of contact.
Exactly 2 tangents. From a point on the circle: 1. From inside: 0.
For triangle with incircle: AP = AS, BP = BQ, CQ = CR (tangent pairs).
A tangent is a line that touches the circle at exactly one point, called the point of tangency or point of contact.
Must-Know Questions
Q1The number of tangents that can be drawn to a circle from a point inside it is:
No tangent can be drawn from a point inside the circle.
Q2A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that BD = 8 cm and DC = 6 cm. Find the sides AB and AC if AB and AC are tangents from B and C.
Let circle touch AB at E, AC at F, BC at D. BD = BE = 8, DC = CF = 6. Let AE = AF = x. AB = x + 8, AC = x + 6, BC = 14. Area = √(s(s-a)(s-b)(s-c)) where s = (x+8+x+6+14)/2 = x+14. Also Area = r×s = 4(x+14). By Heron's: 4(x+14) = √((x+14)(6)(8)(x)). 16(x+14)² = 48x(x+14). 16(x+14) = 48x. x+14 = 3x. x = 7. AB = 15, AC = 13.
Q3Read the following: A circle is inscribed in △ABC with AB = 12 cm, BC = 8 cm, AC = 10 cm. The circle touches AB at P, BC at Q, AC at R. (i) Find AP. (ii) Find BQ. (iii) If the radius is r, express the area of △ABC in terms of r.
Let AP = AR = x, BP = BQ = y, CQ = CR = z. x + y = 12, y + z = 8, x + z = 10. Add all: 2(x+y+z) = 30 → x+y+z = 15. x = 15-8 = 7. y = 15-10 = 5. z = 15-12 = 3. (i) AP = 7. (ii) BQ = 5. (iii) s = 15, Area = r × s = 15r.
Q4If a circle of radius r is inscribed in a triangle of area A and semi-perimeter s, then r = ______.
Area = r × s, so r = A/s (inradius formula).
Q5In the given figure, the quadrilateral ABCD is circumscribed to a circle with centre O. If ∠AOB = 115°, then ∠COD =
For a quadrilateral circumscribed about a circle, opposite angles subtended by opposite sides at the centre are supplementary. ∠AOB + ∠COD = 180°. So ∠COD = 180° − 115° = 65°.
Practice Circles
Reinforce what you just revised with practice questions