Tangent to circle properties — length, angle, power of a point

Question

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 and the angle between the two tangents.

(CBSE Class 10 pattern)

Solution — Step by Step

The most fundamental property: the radius drawn to the point of tangency is perpendicular to the tangent.

So $OA \perp PA$ and $OB \perp PB$ .

flowchart TD
    A["Tangent Properties"] --> B["1. Radius ⊥ Tangent\nat point of contact"]
    A --> C["2. Tangent lengths\nfrom external point\nare equal: PA = PB"]
    A --> D["3. OP bisects angle\nbetween tangents\nand angle AOB"]
    B --> E["Creates right triangle\nOAP with ∠OAP = 90°"]

In right triangle OAP:

PA^2 = OP^2 - OA^2 = 13^2 - 5^2 = 169 - 25 = 144

PA = \mathbf{12 \text{ cm}}

Since tangent lengths from an external point are equal: $PB = PA = 12$ cm.

In triangle OAP:

\tan(\angle OPA) = \frac{OA}{PA} = \frac{5}{12}

\angle OPA = \tan^{-1}\left(\frac{5}{12}\right) \approx 22.6°

Since OP bisects the angle between the two tangents:

\angle APB = 2 \times \angle OPA = 2 \times 22.6° \approx \mathbf{45.2°}

Alternatively, $\angle AOB = 180° - \angle APB$ (since OAPB is a cyclic quadrilateral with $\angle A = \angle B = 90°$ , and the sum of opposite angles in OAPB = 180°).

Why This Works

The tangent-radius perpendicularity creates right triangles, which we can solve using Pythagoras and trigonometry. The equal tangent length property follows from congruence: triangles OAP and OBP are congruent (RHS congruence: OA = OB, OP is common, angle A = angle B = 90°), so PA = PB.

The quadrilateral OAPB has two right angles at A and B, so $\angle AOB + \angle APB = 180°$ . This relationship connects the angle at the centre with the angle between the tangents — a useful shortcut.

Alternative Method — Using the Power of a Point

The power of point P with respect to the circle:

\text{Power} = OP^2 - r^2 = PA^2 = PB^2

If a secant from P intersects the circle at points C and D:

PA^2 = PC \times PD

This “power of a point” theorem connects tangent length with secant lengths — useful in advanced problems.

In CBSE Class 10, tangent problems follow a predictable pattern: draw the figure, mark the right angles, apply Pythagoras. The three properties to remember: (1) radius ⊥ tangent, (2) equal tangent lengths, (3) the line joining the external point to the centre bisects both the angle between tangents and the angle subtended at the centre.

Common Mistake

Students often assume the angle between two tangents from an external point is always 90°. This is only true in the special case where $OP = r\sqrt{2}$ . In general, the angle depends on the ratio $r/OP$ . Always calculate it from the right triangle rather than assuming 90°. Another common error: forgetting that OA is the radius (5 cm), not the diameter.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using the Power of a Point

Common Mistake

Try These Next