Two tangents PA and PB from external point P — prove PA = PB

Question

From an external point P, two tangents PA and PB are drawn to a circle with centre O. Prove that PA = PB.

Solution — Step by Step

Draw circle with centre O. Point P is outside the circle. PA and PB are tangents touching the circle at A and B respectively.

Given: PA and PB are tangents to the circle from external point P.

To prove: PA = PB.

Key fact we’ll use: A tangent to a circle is perpendicular to the radius at the point of tangency.

Draw OA, OB (radii to the points of tangency), and OP (joining centre to external point).

Since PA is a tangent at A: $\angle OAP = 90°$

Since PB is a tangent at B: $\angle OBP = 90°$

This gives us two right triangles: $\triangle OAP$ and $\triangle OBP$ .

In $\triangle OAP$ and $\triangle OBP$ :

$OA = OB$ (radii of the same circle)
$OP = OP$ (common hypotenuse)
$\angle OAP = \angle OBP = 90°$ (tangent $\perp$ radius)

By RHS congruence (Right angle — Hypotenuse — Side): $\triangle OAP \cong \triangle OBP$

Since $\triangle OAP \cong \triangle OBP$ , by CPCT (Corresponding Parts of Congruent Triangles):

\boxed{PA = PB}

Hence proved. The two tangents from an external point to a circle are equal in length.

Why This Works

The proof rests on one geometric truth: a tangent meets the radius at exactly 90°. This isn’t just a rule to memorise — it happens because the tangent line touches the circle at exactly one point, and the shortest distance from the centre to any line is the perpendicular. So OA must be perpendicular to PA.

Once we have two right angles plus the same hypotenuse OP and equal radii, RHS congruence locks in the result. CPCT then gives us what we need.

This result has a beautiful consequence: $\angle OPA = \angle OPB$ (also from CPCT), meaning OP bisects the angle between the two tangents.

Alternative Method — Using Pythagoras

In right $\triangle OAP$ : $PA^2 = OP^2 - OA^2$

In right $\triangle OBP$ : $PB^2 = OP^2 - OB^2$

Since $OA = OB$ (radii): $PA^2 = PB^2$ , so $PA = PB$ (taking positive square root, since lengths are positive).

This approach is faster if you’re in a hurry, but RHS congruence is the standard expected method in CBSE boards.

Common Mistake

Using SSS or SAS instead of RHS. Some students try to prove congruence without first establishing the right angles, jumping to “OA = OB and OP = OP, so…” — but that’s only 2 sides, which isn’t enough. The right angles ( $\angle OAP = \angle OBP = 90°$ ) are essential. Always state them explicitly before applying RHS.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Pythagoras

Common Mistake

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