Circle theorems — tangent, chord, secant angle relationships

Question

What are the key circle theorems involving tangents, chords, and secants, and how do we decide which theorem to apply in a given problem?

(CBSE 10 Board — 3-5 marks question)

Solution — Step by Step

Theorem	Statement
Tangent-Radius	A tangent at any point is perpendicular to the radius at that point
Two Tangents	Tangents drawn from an external point are equal in length
Angle in Semicircle	Angle inscribed in a semicircle is $90°$
Inscribed Angle	Angle subtended at centre = $2 \times$ angle subtended at circumference
Chord-Tangent Angle	Angle between tangent and chord = inscribed angle in the alternate segment
Equal Chords	Equal chords are equidistant from the centre

Problem: From an external point $P$ , two tangents $PA$ and $PB$ are drawn to a circle with centre $O$ and radius 5 cm. If $OP = 13$ cm, find the length of each tangent.

Since tangent $\perp$ radius: $OA \perp PA$ . Triangle $OAP$ is right-angled at $A$ .

PA = \sqrt{OP^2 - OA^2} = \sqrt{169 - 25} = \sqrt{144} = \mathbf{12 \text{ cm}}

By the two-tangent theorem, $PB = PA = 12$ cm as well.

If a tangent at point $A$ makes an angle of $50°$ with chord $AB$ , then the inscribed angle in the alternate segment (angle $ACB$ where $C$ is on the major arc) is also $50°$ .

This theorem is a CBSE favourite. The key word is “alternate segment” — the angle goes to the arc on the other side of the chord from the tangent.

If chord $AB$ subtends $120°$ at the centre, then the angle at any point on the major arc is:

\text{Inscribed angle} = \frac{120°}{2} = \mathbf{60°}

And the angle at any point on the minor arc is $180° - 60° = 120°$ .

flowchart TD
    A["Circle Geometry Problem"] --> B{"What elements are given?"}
    B -- "Tangent + Radius" --> C["Use: Tangent ⊥ Radius"]
    B -- "Two tangents from external point" --> D["Use: Equal Tangents theorem"]
    B -- "Angle at centre + circumference" --> E["Use: Central angle = 2 × Inscribed angle"]
    B -- "Diameter as chord" --> F["Use: Angle in semicircle = 90°"]
    B -- "Tangent + Chord" --> G["Use: Alternate Segment theorem"]
    B -- "Two equal chords" --> H["Use: Equal chords equidistant from centre"]
    C --> I["Apply Pythagoras in right triangle"]
    D --> I

Why This Works

All circle theorems stem from one fundamental property: the radius is constant. The tangent-radius perpendicularity comes from the fact that the tangent touches at exactly one point, and the shortest distance from centre to the tangent line must be the radius. The inscribed angle theorem comes from the relationship between arc length and the angle it subtends.

The two-tangent theorem follows from congruent triangles: triangles $OAP$ and $OBP$ share $OP$ , have equal radii $OA = OB$ , and both have right angles. By RHS congruence, $PA = PB$ .

Alternative Method

For problems with tangent lengths, you can use the property that the tangent from external point $P$ satisfies:

PA^2 = PO^2 - r^2

This is just Pythagoras, but memorising it as a formula saves time in CBSE boards.

In CBSE 10, circle theorem proofs carry 3-5 marks. The two most commonly asked proofs are: (1) tangent is perpendicular to radius, and (2) tangents from an external point are equal. Practise writing these proofs in under 5 minutes.

Common Mistake

Students confuse the angle in the alternate segment with the angle in the same segment. When a tangent at $A$ makes an angle with chord $AB$ , the equal inscribed angle is on the OTHER side of the chord (the alternate segment). Drawing the diagram carefully — marking which segment is “alternate” — prevents this error.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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