Geometry path · G11

Before this lesson: Angle chasing, Similar triangles, Pythagoras, medians and Stewart’s theorem

Your goal: Combine angle and length properties while checking the point’s position.

Start with the idea and one example. Take a break before the written task if you need it. The questions are original teaching exercises, not official past-paper questions.

Names you may know: intersecting chords theorem; tangent–secant theorem; radical axis.

Power of a Point, Circle Chords and Tangents: the key idea

Equal chords of a circle subtend equal central angles, and the perpendicular from the centre to a chord bisects it. An inscribed angle is half the central angle subtending the same arc. A tangent is perpendicular to the radius at contact; the tangent–chord angle equals the angle in the alternate segment. For intersecting chords through P, PA·PB=PC·PD in the usual unsigned interior configuration. For an exterior secant and tangent, PT²=PA·PB, where A and B lie on the same ray from P. The signed quantity OP²−r² is the power of P. Equal powers define a radical axis, a line perpendicular to the line of distinct centres when it is nonempty.

A circle crossed by a secant from exterior point P.PABTO
A diameter secant and a tangent from P. The formula uses PA and the whole length PB. This schematic has different lengths from the numerical example.

A worked example

From an exterior point P, a secant meets the circle at A then B with PA=4 and AB=5. Find tangent length PT.

The whole secant PB=4+5=9. Power gives PT²=PA·PB=4·9=36, so PT=6. Use the full secant, not just the internal chord.

Your turn: change one thing

Two chords AB and CD meet inside at P. If PA=3,PB=8,PC=4, find PD.

Try this on paper before opening the explanation.

Compare your reasoning

The chord products agree: 3·8=4·PD, so PD=6.

Methods and connections

Why chord products agree

For chords AB and CD crossing at P inside the circle, triangles APC and DPB are similar. Their angles at P are vertically opposite, while ∠ACP=∠DBP subtend the same arc AD. Correspondence gives PA/PD=PC/PB, hence PA·PB=PC·PD. State the correspondence before multiplying ratios.

Radical axes and three circles

Subtract the two power expressions |P−O₁|²−r₁² and |P−O₂|²−r₂². The quadratic terms cancel, leaving a linear equation in P. For distinct centres this is their radical axis, perpendicular to O₁O₂. With three circles, if two radical axes meet at P, then P has equal power to all three circles and therefore lies on the third radical axis as well. Parallel-axis and concentric degeneracies must be handled separately.

Common tangent lengths

For disjoint circles with centre distance d and radii R,r, a direct common tangent has between-contact length d2−(R−r)2\sqrt{d^{2}-(R-r)^{2}}. A transverse common tangent has length d2−(R+r)2\sqrt{d^{2}-(R+r)^{2}} when d>R+r. Draw the radii perpendicular to the tangent; their difference or sum becomes one leg of a right triangle. Tangency limits may give zero length or fewer distinct tangent lines.

Pause and check

A trap to avoid: Mixing signed powers with unsigned products or forgetting tangent existence conditions.

Practise and adjust the level

Foundation checks the language; Core applies the method; Stretch asks you to choose or justify an idea. These are levels within this lesson. A session selects six of the nine questions; unused questions allow the level to change. Advanced theory still needs written practice.

Interactive practice loads here. You can also use the complete question set below.

Write a complete argument

Prove equal tangents from an exterior point have equal lengths.

Planning hint

List the assumptions and the exact conclusion. Identify the definition or theorem in this lesson that connects them. Explain why its conditions hold before using it.

Read the full solution after your attempt

Let tangents from P touch the circle at T,U with centre O. Radii OT and OU are perpendicular to the tangents. Right triangles OTP and OUP have common hypotenuse OP and equal radii OT=OU. RHS congruence gives PT=PU.

My proof notebook

Write on paper, or keep a draft here. Compare your reasoning with the solution only after a real attempt. The checklist is your own review, not an automatic mark.

All nine practice questions

Prefer paper or have JavaScript switched off? The complete question set, hints and solutions are here. Interactive practice uses these same questions in an order chosen from your answers.

1. A tangent is perpendicular to which segment at contact T?

Foundation

  1. Every chord through T
  2. Every secant
  3. The diameter not through T
  4. The radius OT
Hint

Join the contact point to the centre.

Answer and reasoning

The radius OT. The radius to the point of tangency is perpendicular to the tangent.

2. An inscribed angle is what fraction of the matching central angle?

Foundation

  1. Equal always
  2. 1/2
  3. 2
  4. 1/3
Hint

Both must subtend the same arc.

Answer and reasoning

1/2. The angle at the circumference is half the angle at the centre.

3. PA=4,AB=5 along an exterior secant P–A–B. What is PB?

Core

  1. 20
  2. 9
  3. 5
  4. 1
Hint

The whole secant includes both segments.

Answer and reasoning

9. PB=PA+AB=9.

4. For that secant, what is the tangent length?

Core

  1. 20\sqrt{20}
  2. 9
  3. 4
  4. 6
Hint

Use PT²=4·9.

Answer and reasoning

6. PT=36=6PT=\sqrt{36}=6.

5. Intersecting chords have PA=3,PB=8,PC=4. What is PD?

Stretch

  1. 24
  2. 6
  3. 2
  4. 12
Hint

Equate chord products.

Answer and reasoning

6. PD=(3·8)/4=6.

6. Why is PA·AB generally wrong in the tangent–secant formula?

Stretch

  1. Only diameters can be secants
  2. A and B coincide
  3. The second factor must be the whole secant PB
  4. Tangents have zero length
Hint

Power uses both distances from the external point.

Answer and reasoning

The second factor must be the whole secant PB. The correct product is PA·PB, with the same starting point P.

7. A chord has both endpoints where?

Foundation

  1. On the circle
  2. At the centre
  3. Outside the circle only
  4. On a tangent line away from contact
Hint

Use the definition of a chord.

Answer and reasoning

On the circle. It joins two points of the circumference.

8. An inscribed angle subtends an arc of 100°. Its measure?

Core

  1. 80°
  2. 50°
  3. 100°
  4. 200°
Hint

Take half the intercepted arc measure.

Answer and reasoning

50°. The inscribed-angle theorem gives 100°/2=50°.

9. For interior intersecting chords, why is the point’s position relevant?

Stretch

  1. Circles have no exterior points
  2. Every distance is negative
  3. The theorem uses only radii
  4. It determines segment order and the form of the product relation
Hint

Interior and exterior configurations use different segment interpretations.

Answer and reasoning

It determines segment order and the form of the product relation. One must identify the two distances from the same intersection point correctly.

Choose your next step

Continue to Cyclic and tangential quadrilaterals. If this felt difficult, return to a prerequisite above. Every lesson stays open.

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