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Tangent-Secant Theorem: PA × PB = PT² | Circles

Learn the tangent–secant theorem (power of a point), its proof using similar triangles, the intersecting chords theorem, and two solved examples. Enrichment beyond Class 10.

5 min read

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scholar 247

Updated 25 October 2026

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Quick answer: If from an external point PP a secant cuts a circle at AA and BB, and a tangent touches it at TT, then PA×PB=PT2PA \times PB = PT^2. This is the tangent–secant theorem (also called the power of a point). For example, if PA=4 cmPA = 4\text{ cm} and AB=5 cmAB = 5\text{ cm}, then PB=9 cmPB = 9\text{ cm} and PT=4×9=6 cmPT = \sqrt{4 \times 9} = 6\text{ cm}. Inside a circle, for two chords ABAB and CDCD meeting at PP: PA×PB=PC×PDPA \times PB = PC \times PD.

From a point outside a circle, a line can either touch the circle at one point (a tangent) or cut it at two points (a secant). The tangent–secant theorem links their lengths with a simple equation. A similar result holds for two chords that cross inside a circle.

Syllabus note: The CBSE Class 10 NCERT chapter Circles covers tangents to a circle (for example, that the tangent is perpendicular to the radius, and that the two tangents from an external point are equal). The tangent–secant theorem, the alternate segment theorem and the intersecting chords theorem are not part of the Class 10 syllabus. This article is enrichment material.


What You Will Learn

  • What tangents and secants are
  • The statement of the tangent–secant theorem, PA×PB=PT2PA \times PB = PT^2
  • A proof using similar triangles
  • The intersecting chords theorem, PA×PB=PC×PDPA \times PB = PC \times PD
  • Two solved examples

1. The Theorem

Tangent–secant theorem: If a secant from an external point PP meets a circle at AA and BB (with AA nearer to PP), and a tangent from PP touches the circle at TT, then PA×PB=PT2PA \times PB = PT^2

          T
         /
        /  (circle through T, A, B)
       /
  P --A--------B
  PA = outer part of the secant
  PB = whole secant (P to B)
  PT = tangent
LengthMeaning
PAPAFrom PP to the nearer point where the secant meets the circle
PBPBFrom PP to the farther point, i.e. the whole secant
PTPTThe tangent from PP to the point of contact TT

2. Proof Using Similar Triangles

Construction: Join TATA and TBTB.

Step 1: Use the alternate segment theorem. The angle between a tangent and a chord at the point of contact equals the angle in the alternate segment. So:

∠PTA=∠PBT\angle PTA = \angle PBT

Step 2: Compare triangles PATPAT and PTBPTB.

  • ∠TPA=∠BPT\angle TPA = \angle BPT (common angle at PP)
  • ∠PTA=∠PBT\angle PTA = \angle PBT (from Step 1)

So, by the AA similarity criterion, △PAT∼△PTB\triangle PAT \sim \triangle PTB.

Step 3: Write the ratio of corresponding sides.

PAPT=PTPB\frac{PA}{PT} = \frac{PT}{PB}

Step 4: Cross-multiply.

PA×PB=PT2PA \times PB = PT^2

Hence proved.


3. The Intersecting Chords Theorem

Intersecting chords theorem: If two chords ABAB and CDCD of a circle meet at a point PP inside the circle, then PA×PB=PC×PDPA \times PB = PC \times PD

Proof outline: Join ACAC and BDBD. In triangles PACPAC and PDBPDB:

  • ∠APC=∠DPB\angle APC = \angle DPB (vertically opposite angles)
  • ∠PAC=∠PDB\angle PAC = \angle PDB (angles in the same segment, both standing on arc BCBC)

So △PAC∼△PDB\triangle PAC \sim \triangle PDB, which gives PAPD=PCPB\frac{PA}{PD} = \frac{PC}{PB}, i.e. PA×PB=PC×PDPA \times PB = PC \times PD.


4. Example 1: Finding the Tangent

Question

From an external point PP, a secant PABPAB meets a circle at AA and BB. If PA=4 cmPA = 4\text{ cm} and AB=5 cmAB = 5\text{ cm}, find the length of the tangent PTPT.

Solution

Step 1: Find the whole secant.

PB=PA+AB=4+5=9 cmPB = PA + AB = 4 + 5 = 9\text{ cm}

Step 2: Apply the theorem.

PT2=PA×PB=4×9=36  ⟹  PT=6 cmPT^2 = PA \times PB = 4 \times 9 = 36 \implies PT = 6\text{ cm}

Answer: PT=6 cmPT = 6\text{ cm}.


5. Example 2: Intersecting Chords

Question

Chords ABAB and CDCD of a circle meet at an interior point PP. If PA=3 cmPA = 3\text{ cm}, PB=8 cmPB = 8\text{ cm} and PC=4 cmPC = 4\text{ cm}, find PDPD and the length of CDCD.

Solution

Step 1: Apply the theorem.

PA×PB=PC×PD  ⟹  3×8=4×PD  ⟹  PD=6 cmPA \times PB = PC \times PD \implies 3 \times 8 = 4 \times PD \implies PD = 6\text{ cm}

Step 2: Find CDCD.

CD=PC+PD=4+6=10 cmCD = PC + PD = 4 + 6 = 10\text{ cm}

Answer: PD=6 cmPD = 6\text{ cm} and CD=10 cmCD = 10\text{ cm}.


6. Summary Table

SituationResultSimilar triangles used
Tangent and secant from an external pointPA×PB=PT2PA \times PB = PT^2△PAT∼△PTB\triangle PAT \sim \triangle PTB
Two chords meeting inside the circlePA×PB=PC×PDPA \times PB = PC \times PD△PAC∼△PDB\triangle PAC \sim \triangle PDB
Two secants from an external pointPA×PB=PC×PDPA \times PB = PC \times PDBoth products equal PT2PT^2

7. Tips and Common Mistakes

Tip: PBPB is the whole secant, from PP to the far point BB: PB=PA+ABPB = PA + AB.

Common Mistake: Writing PA×AB=PT2PA \times AB = PT^2. The correct statement is PA×PB=PT2PA \times PB = PT^2.

Common Mistake: Using the theorem for a point inside the circle. Inside the circle, use the intersecting chords form, PA×PB=PC×PDPA \times PB = PC \times PD.


Frequently Asked Questions

What is the tangent–secant theorem?

If a secant from an external point P meets a circle at A and B, and a tangent from P touches it at T, then PA × PB = PT².

How do you find the length of a tangent using a secant?

Find the whole secant PB = PA + AB, then use PT = √(PA × PB).

What is the intersecting chords theorem?

If two chords AB and CD meet at a point P inside a circle, then PA × PB = PC × PD.

Is the tangent–secant theorem in the Class 10 syllabus?

No. It is not part of the current CBSE Class 10 syllabus; it is enrichment material.

Concept Check

MEDIUM

For what value of kk will the following pair of linear equations have NO solution? 3x+y=13x + y = 1 (2k−1)x+(k−1)y=2k+1(2k - 1)x + (k - 1)y = 2k + 1

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