Prove That the Intercept of a Tangent Between Two Parallel Tangents to a Circle Subtends a Right Angle at the Centre
Prove that the intercept of a tangent between two parallel tangents to a circle subtends a right angle at the centre.
Given: XY and X′Y′ at are two parallel tangents to the circle with centre O, and AB is the tangent at the point C, which intersects XY at A and X′Y′ at B.
To Prove: ∠AOB = 90°
Construction: Let us joint point O to C.
Proof:
In ΔOPA and ΔOCA, we have
OP = OC (Radii of the same circle)
AP = AC (Tangents from point A)
AO = AO (Common side)
ΔOPA ≅ ΔOCA (SSS congruence criterion)
Therefore, ∠POA = ∠COA ……(i) (C.P.C.T)
Similarly, ΔOQB ≅ ΔOCB ……(ii)
Since POQ is a diameter of the circle, it is a straight line.
Therefore, ∠POA + ∠COA + ∠COB + ∠QOB = 180°
From equations (i) and (ii), it can be observed that
2∠COA + 2∠COB = 180°
∴ ∠COA + ∠COB = 90°
So, ∠AOB = 90°.
Frequently Asked Questions
Find answers to common questions.
Why are the tangents from an external point equal in length?
Tangents from an external point to a circle are equal because the two right-angled triangles formed (by the radius to each tangent point and the line from the external point to the centre) are congruent by the RHS criterion — same hypotenuse (distance to centre) and same leg (radius).
Which congruence criterion is used in this proof?
The SSS (Side-Side-Side) congruence criterion is used twice in this proof — first for triangles OPA and OCA, and then for triangles OQB and OCB. The equal sides come from radii of the same circle, tangent-equality from an external point, and a shared side (AO or OB).
What does it mean for a tangent intercept to subtend a right angle at the centre?
It means the angle ∠AOB, formed at the centre O between the two lines OA and OB (where A and B are the points where the third tangent meets the two parallel tangents), is exactly 90°. This holds true regardless of where on the circle the third tangent touches.
What happens to ∠AOB if the third tangent is drawn at different positions on the circle?
∠AOB always equals 90°, regardless of where on the circle the third tangent is drawn. This is the power of the theorem — the right angle at the centre is an invariant property of this configuration for any circle with two fixed parallel tangents.
Can this result be proved using any other method?
Yes. An alternative approach uses the angle-bisector property more directly: since OA bisects ∠POC and OB bisects ∠QOC (both derived from tangent-equality), and ∠POC + ∠QOC = 180° (straight line), it follows that (½)∠POC + (½)∠QOC = 90°, giving ∠AOB = 90°. The SSS method shown above is the standard NCERT method
Is this theorem part of the CBSE Class 10 syllabus?
Yes. This theorem is part of NCERT Class 10 Mathematics, Chapter 10 — Circles. It is a proof-type question and has appeared in CBSE board exams as a 3 or 4-mark question. Students aiming for full marks should be able to reproduce the proof with correct diagram, steps, and reasons.