Ever wondered why a straight line can just kiss a circle at one spot and never cross it? That line is called a tangent, and it hides some neat tricks that show up a lot in your CBSE exams.
A tangent is a line that touches a circle at exactly one point. The magic? The radius drawn to that point always meets the tangent at a right angle. Think of a road that just grazes a round lake without diving in.
What is a Tangent?
A tangent (pronounced tan-jent) is a straight line that meets a circle in only one place. That single meeting spot is called the point of contact. If you draw the radius – the line from the circle’s centre to any point on the circle – to the point of contact, the radius and the tangent form a 90° angle, just like the corner of a sheet of paper.
Why the radius is perpendicular
Imagine the circle as a round pond and the radius as a stick poking straight down from the centre to the water’s surface. If you slide a flat board (the tangent) so it just touches the stick’s tip, the board will be perfectly flat against the stick – that’s a right angle.
Main Theorems About Tangents
Theorem 1 – Tangent at a point is perpendicular to the radius
Statement: The line drawn from the centre of a circle to the point where a tangent touches the circle is perpendicular to the tangent.
Proof (quick version): Let O be the centre, P the point of contact, and T the tangent line. Suppose OT is not a right angle. Then you could draw a shorter line from O to T that still touches the circle, contradicting the definition of a radius as the shortest distance from the centre to the circle. Hence, OT must be 90°.
Example: In a circle with centre (0,0) and radius 5, the line x = 5 is a tangent at point (5,0). The radius to (5,0) is the segment from (0,0) to (5,0) – a horizontal line. The tangent x = 5 is vertical, and horizontal meets vertical at a right angle.
Theorem 2 – Two tangents from an external point are equal
Statement: If you pick a point outside the circle and draw two tangents to the circle, the lengths of those two tangent segments are the same.
Proof: Let external point be A, and the two points of contact be B and C. Join A to the centre O, and also draw OB and OC. Triangles △ABO and △ACO share side AO, have OB = OC (both radii), and have right angles at B and C (by Theorem 1). Two right‑angled triangles with a common hypotenuse and another equal side must be congruent (by the RHS criterion). Therefore AB = AC.
Example: A point P(8,0) lies outside a circle centred at O(0,0) with radius 5. The tangents from P touch the circle at Q and R. Using the distance formula, both PQ and PR turn out to be √(8²‑5²)=√39, confirming the theorem.
Theorem 3 – Angle between tangent and chord equals angle in alternate segment
Statement: The angle formed by a tangent and a chord through the point of contact equals the angle in the opposite part of the circle that subtends the same chord.
Proof sketch: Let AB be a chord, and AT be the tangent at A. Draw the centre O and connect O to B. Triangle OAB is isosceles (OA = OB). The exterior angle at A (∠BAT) equals the interior opposite angle ∠ABO (by the exterior angle theorem). Since ∠ABO equals the angle subtended by chord AB in the alternate segment, the statement follows.
Example: In a circle, chord CD subtends 40° at the centre. If a tangent touches the circle at C, then ∠(tangent‑CD) also measures 40°.
Quick Summary Table
| Theorem | Key Idea | Typical Use in Exams |
|---|---|---|
| Radius ⟂ Tangent | Radius meets tangent at 90° | Identify right‑angle triangles |
| Equal Tangents from External Point | Two tangents from same outside point have equal length | Find unknown lengths |
| Angle between Tangent & Chord | Equals angle in opposite segment standing on same chord | Calculate angles without trigonometry |
Worked Example Combining Theorems
Problem: From point P(12,0) draw tangents to a circle centred at O(0,0) with radius 5. Find the angle between the two tangents.
- Use Theorem 2: Both tangents have equal length. Let the length be t. By the Pythagorean theorem in right triangle OPA, t² = OP² – OA² = 12² – 5² = 144 – 25 = 119, so t = √119.
- Join O to P. Triangle formed by the two radii to the points of contact and OP is isosceles. The angle at O (∠BOC) can be found using cosine rule: cos ∠BOC = (5²+5²‑(2t)²)/(2·5·5). Plugging t = √119 gives cos ∠BOC = (50‑476)/(50) = -426/50 = -8.52, which is impossible, meaning we should instead use geometry: the two tangents are symmetric about OP, so the angle between them is 2·θ where sin θ = radius/OP = 5/12. Hence θ = arcsin(5/12) ≈ 24.6°, so total angle ≈ 49.2°.
This uses Theorem 1 (right angle at each point of contact) and Theorem 2 (equal lengths) to turn a messy algebra problem into a neat trigonometric one.
📝 Likely Exam Questions
- State and prove the theorem that two tangents drawn from an external point to a circle are equal.
Answer: State as above, then draw the figure, join centre, show two right‑angled triangles share a hypotenuse and a side, invoke RHS congruence, conclude equality. - Given a circle with centre O and radius 7 cm, a tangent touches the circle at A. If OA = 7 cm and the tangent makes a 90° angle with OA, find the distance from the centre to any point on the tangent line.
Answer: Any point on the tangent line is at least 7 cm from O, because the shortest distance from O to the line is the radius itself, i.e., 7 cm. - In a circle, a chord AB subtends 50° at the centre. Find the angle between the tangent at A and the chord AB.
Answer: By Theorem 3, the angle equals the angle in the alternate segment, which is half the central angle: 50°/2 = 25°. - From point P outside a circle, two tangents PA and PB are drawn. If PA = 10 cm and the distance OP = 26 cm, find the radius of the circle.
Answer: In right triangle OAP, OA² + PA² = OP² → r² + 10² = 26² → r² = 676 – 100 = 576 → r = 24 cm. - Prove that the angle between a tangent and a chord through the point of contact equals the angle in the alternate segment.
Answer: Provide the construction, use the exterior angle theorem and the fact that equal chords subtend equal angles, concluding the required equality.