⭕ Circles — Class 10

Tangents, number of tangents from a point, and theorems with proofs

1. Introduction — Tangent to a Circle

📖 Key Definitions

Secant: A line that intersects a circle at two points.

Tangent: A line that touches a circle at exactly one point. This point is called the point of tangency or point of contact.

Common tangent: A line that is tangent to two circles simultaneously.

🌟 Real-Life Understanding

A wheel on a road — the road is tangent to the wheel. The wheel touches the road at exactly one point (the contact point) at any given moment. Similarly, a tangent line touches a circle at just one point!

Type of Tangents	Description	Number
External Common	Don't pass between circles	2 (for external circles)
Internal Common	Pass between two circles	2 (for external circles)
From External Point	Both touching same circle	2

2. Theorem 1 — Tangent ⊥ Radius

⚡ Theorem Statement

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

If OP is radius and PQ is tangent at P, then OP ⊥ PQ (∠OPQ = 90°)

📖 Proof (by contradiction)

Given: Circle with centre O, tangent PQ at point P.

To prove: OP ⊥ PQ

Proof: Assume OP is not perpendicular to PQ. Then there exists a point R on PQ such that OR ⊥ PQ (drop a perpendicular). In right triangle ORP: OR < OP (hypotenuse > perpendicular). But OR < OP means R is inside the circle. Since PQ is tangent, it should touch at only one point P. But if R is inside the circle, line PQ would intersect the circle at two points — contradiction! Hence OP ⊥ PQ. ∎

3. Number of Tangents From a Point

Position of Point	Number of Tangents	Reason
Inside the circle	0	No line from inside can touch without crossing
On the circle	1	Only one tangent at that point
Outside the circle	2	Two tangents can be drawn

4. Theorem 2 — Equal Tangents from External Point

⚡ Theorem Statement

The lengths of two tangents drawn from an external point to a circle are equal.

If PA and PB are tangents from external point P to circle with centre O, then PA = PB

📖 Proof

Given: PA and PB are tangents from P. O is centre. OA and OB are radii.

To prove: PA = PB

Proof:

In △OAP and △OBP:

OA = OB (radii of same circle)

OP = OP (common side)

∠OAP = ∠OBP = 90° (radius ⊥ tangent, Theorem 1)

By RHS congruence: △OAP ≅ △OBP

Therefore PA = PB (CPCT) ∎

🔑 Additional Results from Theorem 2

OP bisects angle APB (∠APO = ∠BPO)
OP bisects angle AOB (∠AOP = ∠BOP)
OP is the perpendicular bisector of AB

💡 Example 1: Find length of tangent

Q: A tangent PQ is drawn from external point P to a circle with centre O, radius 5 cm. If OP = 13 cm, find length of PQ.

Since OQ ⊥ PQ: OQ² + PQ² = OP² (Pythagoras)

5² + PQ² = 13² → 25 + PQ² = 169 → PQ² = 144 → PQ = 12 cm

💡 Example 2: Perimeter of quadrilateral

Q: A circle touches all four sides of quadrilateral ABCD. Prove that AB + CD = BC + DA.

Let P, Q, R, S be points of contact on AB, BC, CD, DA respectively.

Since tangents from external point are equal:

From A: AP = AS | From B: BP = BQ | From C: CQ = CR | From D: DR = DS

AB + CD = (AP + PB) + (CR + RD) = (AS + BQ) + (CQ + DS)

= (AS + DS) + (BQ + CQ) = DA + BC = BC + DA ✓

💡 Example 3: Angle at centre

Q: From a point P, two tangents PA and PB are drawn to a circle with centre O, radius 3 cm. If PA = 4 cm, find ∠APB.

OP = √(OA² + PA²) = √(9 + 16) = 5 cm

In △OAP: tanα = OA/PA = 3/4, where α = ∠APO

α = tan⁻¹(3/4) ≈ 36.87°

∠APB = 2α ≈ 73.74°

Alternatively: sin∠APO = OA/OP = 3/5 → ∠APO = 37° → ∠APB ≈ 74°

5. Summary of Key Points

Tangent meets circle at exactly one point.
Tangent is perpendicular to radius at point of contact.
From an external point, two tangents of equal length can be drawn.
No tangent can be drawn from a point inside the circle.
OAP is always a right angle (radius to tangent point).

📋 Key Results — Circles

Radius ⊥ Tangent

∠OPQ = 90°

(O = centre, P = contact point)

Equal Tangents

PA = PB

(P = external point)

Length of Tangent

PT = √(OP² – r²)

(r = radius, OP = dist from centre)