Circles for CDS

Circle questions in CDS look harder than they are. Nearly every one rests on a small set of theorems: the angle at the centre, angles in the same segment, the cyclic quadrilateral, and the tangent meeting the radius at a right angle. Add Pythagoras and you can handle most of what the paper asks.

The skill lies in reading the figure. Most questions describe it in words: "PA and PB are tangents", "ABCD is cyclic", "AB is a diameter". Each phrase is a signal for one theorem. This guide lists those signals, explains why each theorem holds, and then works through CDS-style questions.

Chords

• The perpendicular from the centre to a chord bisects the chord. So if a chord of length c is at distance d from the centre of a circle of radius r, then d² + (c/2)² = r².
• Equal chords are equidistant from the centre, and equal chords subtend equal angles at the centre.
• A longer chord is nearer the centre. The diameter, the longest chord, passes through it.
• A chord equal in length to the radius subtends 60° at the centre, because it forms an equilateral triangle with the two radii.

Angles in a circle

• Angle at the centre: the angle an arc subtends at the centre is twice the angle it subtends at any point on the remaining circle. Why: draw the radius to that point and use the exterior angle of the two isosceles triangles it creates.
• Same segment: angles subtended by the same arc at points in the same segment are equal.
• Semicircle: the angle in a semicircle is 90°. This is the centre theorem with a 180° central angle.
• Cyclic quadrilateral: opposite angles add to 180°. The two opposite angles stand on arcs that together make the whole circle, so their central angles add to 360°.
• Exterior angle of a cyclic quadrilateral equals the interior opposite angle.

Tangents

• A tangent is perpendicular to the radius at the point of contact.
• The two tangents from an external point are equal in length. The line from that point to the centre bisects the angle between them.
• Tangent length from a point at distance d from the centre: √(d² − r²).
• If PA and PB are tangents from P and O is the centre, then ∠APB + ∠AOB = 180°, because quadrilateral OAPB has two right angles.
• Alternate segment theorem: the angle between a tangent and a chord through the point of contact equals the angle the chord subtends in the alternate segment.
• If a quadrilateral is drawn around a circle, touching all four sides, then AB + CD = AD + BC. This follows from equal tangents at each corner.

Products of lengths

• Intersecting chords: if chords AB and CD cut at P inside the circle, then AP × PB = CP × PD.
• Tangent and secant: if PT is a tangent and a line from P cuts the circle at A and B, then PT² = PA × PB.

Both come from similar triangles formed by angles in the same segment.

Two circles

Let the centres be d apart and the radii r₁ and r₂.

• Length of a direct common tangent: √(d² − (r₁ − r₂)²)
• Length of a transverse common tangent: √(d² − (r₁ + r₂)²)

Worked questions

Question 1: ABCD is a cyclic quadrilateral with ∠A = (2x + 10)° and ∠C = (3x − 5)°. Also ∠B = 70°. Find all four angles.

• ∠A + ∠C = 180°: 5x + 5 = 180, so x = 35.
• ∠A = 80°, ∠C = 100°, ∠B = 70°, ∠D = 110°.

Question 2: Two parallel chords of lengths 6 cm and 8 cm lie in a circle of radius 5 cm. Find the distance between them if they are on opposite sides of the centre, and if they are on the same side.

• Distance of the 6 cm chord from the centre: √(25 − 9) = 4 cm. Distance of the 8 cm chord: √(25 − 16) = 3 cm.
• Opposite sides: 4 + 3 = 7 cm. Same side: 4 − 3 = 1 cm.

Question 3: PA and PB are tangents from P to a circle with centre O, and ∠APB = 50°. Find ∠AOB, and the angle that chord AB subtends at a point C on the major arc.

• ∠AOB = 180° − 50° = 130°.
• ∠ACB = half of 130° = 65°.

Question 4: Chords AB and CD intersect at P inside a circle. AP = 6 cm, PB = 4 cm and CP = 3 cm. Find PD.

• 6 × 4 = 3 × PD, so PD = 8 cm.

Question 5: From a point P, a tangent PT and a line cutting the circle at A and B are drawn, with A between P and B. PA = 4 cm and AB = 5 cm. Find PT.

• PB = 4 + 5 = 9 cm. PT² = 4 × 9 = 36, so PT = 6 cm.

Question 6: Two circles of radii 8 cm and 3 cm have their centres 13 cm apart. Find the lengths of the direct and transverse common tangents.

• Direct: √(169 − 25) = √144 = 12 cm.
• Transverse: √(169 − 121) = √48 = 4√3 cm, about 6.9 cm.

Practice set

1. What is the angle in a semicircle?
2. A chord 24 cm long lies in a circle of radius 13 cm. How far is it from the centre?
3. A point is 25 cm from the centre of a circle of radius 7 cm. Find the length of the tangent from it.
4. Quadrilateral ABCD is drawn around a circle, touching all four sides. AB = 6 cm, BC = 7 cm and CD = 4 cm. Find AD.
5. The tangent at A makes an angle of 40° with chord AB. Find the angle AB subtends at a point C in the alternate segment.
6. ABCD is cyclic, and side AB is produced to E. If ∠CBE = 70°, find ∠ADC.
7. Two circles of radii 5 cm and 3 cm touch externally. Find the distance between their centres and the number of common tangents.
8. A chord equal to the radius subtends what angle at a point on the major arc?

Answers

1. 90°.
2. 5 cm. √(169 − 144).
3. 24 cm. √(625 − 49) = √576.
4. 3 cm. AB + CD = AD + BC, so 10 = AD + 7.
5. 40°. Alternate segment theorem.
6. 70°. The exterior angle equals the interior opposite angle.
7. 8 cm and 3 tangents.
8. 30°. The chord subtends 60° at the centre, and half of that at the circumference.

What to do next

• Write the list of "signal phrases" (tangent, cyclic, diameter, chords cutting) with the theorem each one triggers
• Redraw Questions 3 and 5 from their wording alone
• Revise lines, angles and triangles, because circle proofs lean on isosceles and similar triangles
• Move on to area and perimeter for sectors, segments and rings
• Solve the circle questions from the last five CDS papers under time