Tangent–Secant Power

\( t^{2}=e(\,e+\ell\,) \)
Circle Theorems GCSE

\( e=4, ℓ=10. Find tangent length. \)

Tips: use ^ for powers, sqrt() for roots, and type pi for π.
Hint (H)
\( t²=4(14) \)

Explanation

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Statement

The tangent–secant power theorem relates the length of a tangent from a point outside a circle to the lengths of a secant drawn from the same point. If a tangent of length \(t\) touches the circle, and a secant passes through the circle with external part length \(e\) and internal part length \(\ell\), then:

\[ t^2 = e(e + \ell) \]

Why it’s true

  • The theorem follows from similar triangles formed between the tangent and the secant inside the circle.
  • When the tangent and secant are drawn from the same external point, the geometry ensures proportional sides, leading to the power equation.
  • This property is part of the more general “power of a point” theorem.

Recipe (how to use it)

  1. Identify tangent length \(t\) if given.
  2. Identify the external length \(e\) of the secant and its internal length \(\ell\).
  3. Apply the formula \(t^2 = e(e+\ell)\).
  4. Solve for the unknown (can be \(t\), \(e\), or \(\ell\)).

Spotting it

This theorem applies whenever a tangent and a secant are drawn from the same external point to a circle. Look for a tangent touching at one point and a secant cutting through at two points.

Common pairings

  • Two–secant power theorem: \(e_1(e_1+\ell_1) = e_2(e_2+\ell_2)\).
  • Tangent–chord angle theorems.

Mini examples

  1. Given: \(e=4\), \(\ell=6\). Find: \(t\). Answer: \(t^2=4(10)=40\), so \(t=\sqrt{40}\approx 6.32\).
  2. Given: \(t=12\), \(e=9\). Find: \(\ell\). Answer: \(144=9(9+\ell)\), \(\ell=7\).

Pitfalls

  • Forgetting that \(e+\ell\) is the full secant length.
  • Mixing up tangent length with secant parts.
  • Leaving answers unsimplified when square roots are exact.

Exam strategy

  • Always write down the formula first: \(t^2=e(e+\ell)\).
  • Check if the question wants exact surd form (\(\sqrt{}\)) or rounded decimals.
  • Be careful when rearranging equations to solve for missing lengths.

Summary

The tangent–secant power theorem is a powerful circle result: \(t^2=e(e+\ell)\). It links a tangent’s length to the two parts of a secant. Spot it in exam problems whenever you see a tangent and a secant from the same point outside a circle.

Worked examples

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  1. \( A tangent has length t, and a secant with e=4, l=6. Find t. \)
    1. \( t²=e(e+ℓ)=4(10)=40 \)
    2. \( t=√40≈6.32 \)
    Answer: \( \sqrt{40}\,(≈6.32) \)
  2. \( Given tangent length t=12, external secant e=9. Find internal ℓ. \)
    1. \( t²=144 \)
    2. \( 144=9(9+ℓ) \)
    3. \( ℓ=7 \)
    Answer: 7
  3. \( Find tangent length when e=5, ℓ=11. \)
    1. \( t²=5(16)=80 \)
    2. \( t=√80≈8.94 \)
    Answer: \( \sqrt{80}\,(≈8.94) \)
  4. \( Tangent t=10, external e=8. Find ℓ. \)
    1. \( 100=8(8+ℓ) \)
    2. \( 100=64+8ℓ \)
    3. \( ℓ=4.5 \)
    Answer: 4.5
  5. \( e=7, ℓ=9. Find tangent length t. \)
    1. \( t²=7(16)=112 \)
    2. t≈10.58
    Answer: \( \sqrt{112}\,(≈10.58) \)
  6. \( Tangent=15, external=9. Find ℓ. \)
    1. \( 225=9(9+ℓ) \)
    2. \( 225=81+9ℓ \)
    3. \( ℓ=16 \)
    Answer: 16
  7. \( Secant has e=12, ℓ=20. Find t. \)
    1. \( t²=12(32)=384 \)
    2. t≈19.60
    Answer: \( \sqrt{384}\,(≈19.60) \)
  8. \( Tangent=20, external=16. Find ℓ. \)
    1. \( 400=16(16+ℓ) \)
    2. \( 400=256+16ℓ \)
    3. \( ℓ=9 \)
    Answer: 9
  9. \( e=25, ℓ=7. Find tangent t. \)
    1. \( t²=25(32)=800 \)
    2. t≈28.28
    Answer: \( \sqrt{800}\,(≈28.28) \)
  10. \( Tangent=30, external=18. Find ℓ. \)
    1. \( 900=18(18+ℓ) \)
    2. \( 900=324+18ℓ \)
    3. \( ℓ=32 \)
    Answer: 32