Geometry II · Circles & Quadrilaterals

Start with the most basic picture of the chapter. Take a chord AB and drop a perpendicular OM from the centre O to it. Three things happen together:

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So every radius–chord–distance question is one Pythagoras. If any two of r, d and the chord are given, the third comes out: chord = 2√r² − d², d = √r² − (chord2)², r = √d² + (chord2)².

Radhika's circle has radius 41 cm and a chord 9 cm from the centre. Half chord = √1681 − 81 = √1600 = 40, so the chord is 80 cm. The (9, 40, 41) triplet was sitting there all along.

The nearer a chord is to the centre, the longer it is; the diameter (distance 0) is the longest chord. So a shorter chord is always farther from the centre.

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Two parallel chords. Find the distance of each chord from the centre separately. If both chords are on the same side of the centre, the gap between them is the difference d₁ − d₂; if they are on opposite sides, it is the sum d₁ + d₂. When the side is not mentioned, both answers are possible.

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When the radius is unknown and only the gap between two chords is given, write r² from both chords and equate them: (half₁)² + d₁² = (half₂)² + d₂². The gap gives you the second equation.

Concentric circles. If a chord of the bigger circle touches the smaller circle, the point of contact is the foot of the perpendicular from the centre. The chord is bisected there, and its length is 2√R² − r².

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Equal chords from one point. Let A lie on the circle with centre O, and let chords AB and AC be equal. Equal chords are equidistant from the centre, so O is equally far from the lines AB and AC and lies on the bisector of ∠BAC. Hence AO bisects ∠BAC. If ∠BAC = 70°, then ∠OAB = ∠OAC = 35°.

Now the theorem that the whole chapter leans on. Take a chord (or arc) AB. It makes an angle ∠AOB at the centre and an angle ∠ACB at any point C on the remaining part of the circle.

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Why is it true? Join CO and extend it. OA = OC, so triangle OAC is isosceles, and the exterior angle at O on the extended line is ∠OAC + ∠OCA = 2∠OCA. Do the same with triangle OBC and add. The angle at O comes out exactly twice the angle at C.

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Everything else in this type follows from that one theorem:

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Chord from the central angle. Triangle AOB is isosceles with two sides r. If ∠AOB = 60°, it is equilateral and the chord = r. If ∠AOB = 90°, chord = r√2. If ∠AOB = 120°, chord = r√3. If 180°, the chord is the diameter 2r. (In general, chord = 2r sinθ2; you will meet sine in Trigonometry.)

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The isosceles step. In triangle OAB, ∠OAB = ∠OBA = 180° − ∠AOB2. So if C is on the major arc, ∠OAB = 90° − ∠ACB. Many "find ∠OAB" or "find ∠ACB" questions end in one line with this.

Radhika stands at the centre of a circular park and sees two gates A and B at 100°. Her friend on the far edge of the park sees the same two gates at 50°. If he walks along the edge to the near side, between the gates, he sees them at 180° − 50° = 130°.

When the angle bisector reaches the circle. Draw the bisector of ∠A of a triangle ABC and extend it until it meets the circumcircle again at D. Since ∠BAD = ∠DAC, the chords BD and DC subtend equal angles at A, so D is the midpoint of the arc BC (the arc not containing A), and DB = DC. The same equal angles give ∠DBC = ∠DCB = A2, and the line from the centre to D is the perpendicular bisector of BC.

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One more result is worth carrying into the exam. If I is the incentre, then DB = DC = DI. The reason is short: ∠DBI = A2 + B2, and ∠DIB, being the exterior angle of △ABI, is also A2 + B2. So △DBI is isosceles. Radhika's triangle has ∠A = 80°: the bisector meets the circle at D with ∠DBC = ∠DCB = 40° and ∠BDC = 100°.

Chord from the inscribed angle. If a chord subtends an angle θ at a point on the major arc, the angle at the centre is 2θ, and the perpendicular from the centre splits it into two halves of θ. So half the chord is R sin θ, and chord = 2R sin θ. Take R = 10 cm: a 30° angle gives a chord of 10 cm, 45° gives 10√2 ≈ 14.14 cm and 60° gives 10√3 ≈ 17.32 cm. In a triangle ABC this reads BC = 2R sin A. Use θ as the angle on the major arc; for an angle on the minor arc, use 180° − θ.

When two chords cross each other, or two secants (chords produced) meet outside the circle, the angle between them is again decided by arcs. You need only two results, and both come from the exterior-angle property.

Case 1: chords meet inside at P. Chords AB and CD cut at P. Join BC. In triangle PBC, the exterior angle ∠APC = ∠PBC + ∠PCB = ∠ABC + ∠BCD. Now ∠ABC stands on arc AC and ∠BCD stands on arc BD, so each is half of its central angle:

[[fig:0]]

Case 2: secants meet outside at P. Let the lines through P cut the circle at A, B (A nearer) and C, D (C nearer). Join AD. In triangle PAD, the exterior angle ∠BAD = ∠P + ∠ADC, so ∠P = ∠BAD − ∠ADC:

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The figure you will meet most. If the far arc is a semicircle (A and B are the ends of a diameter in the outside case), the formula becomes ∠P = 90° − 12(near central angle). You do not even need the formula. Join a chord to make an angle in a semicircle (90°), and one triangle gives the rest. This diameter-with-secants figure keeps coming back.

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Take Radhika's two crossing chords. The arcs they hold are 100° and 40°, so the inside angle = 12(100 + 40) = 70°. If the same arcs were the far and near arcs of two secants meeting outside, the angle would be 12(100 − 40) = 30°.

A quadrilateral whose four vertices all lie on a circle is a cyclic quadrilateral. Its main property comes straight from Type 2.

In cyclic ABCD, ∠B stands on arc ADC and ∠D stands on arc ABC. These two arcs together make the whole circle, 360° at the centre. So ∠B + ∠D = 12 × 360° = 180°. In the same way ∠A + ∠C = 180°.

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Exterior angle. Extend BC to E. ∠DCE = 180° − ∠BCD = ∠A. So the exterior angle of a cyclic quadrilateral equals the interior opposite angle.

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Diagonals give same-segment pairs. In cyclic ABCD, ∠ABD = ∠ACD (both on arc AD), ∠BAC = ∠BDC (both on arc BC), and so on. Most questions of the "find ∠PCD" kind are solved by spotting two such equal angles.

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Angle at the centre. If O is the centre, the angle that arc ADC makes at O equals 2∠ABC, and the angle that arc ABC makes at O equals 2∠ADC. Together they make 360°.

Special quadrilaterals. A rectangle and a square are always cyclic. A cyclic parallelogram must be a rectangle, because its opposite angles are equal and also supplementary, so each is 90°. A cyclic trapezium is always isosceles. A rhombus is cyclic only when it is a square.

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Sides produced to meet. Suppose AB and DC produced beyond B and C meet at E, and AD and BC produced beyond D and C meet at F, so that C is the vertex nearest to both. Write the angle sums of the two outer triangles ADE and ABF. Adding them and using ∠B + ∠D = 180° gives 2∠A + ∠E + ∠F = 180°.

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Radhika draws a cyclic quadrilateral with ∠A = 75°. Without measuring anything she knows ∠C = 105°, and if she extends BC beyond C, the exterior angle there is also 75°.

(For the curious: Ptolemy's theorem says that in a cyclic quadrilateral AC × BD = AB × CD + AD × BC. SSC rarely needs it, but you should recognise it.)

Ptolemy's theorem and Brahmagupta's formula. Two more results about cyclic quadrilaterals appear now and then, mostly in Mains-level papers. For a cyclic quadrilateral ABCD:

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A rectangle is the easiest check of Ptolemy. With sides 5 and 12 the diagonals are 13 each, and 13 × 13 = 5 × 5 + 12 × 12. So Pythagoras is Ptolemy's theorem for a rectangle. Ptolemy also gives a neat result for an equilateral triangle: if P lies on the minor arc BC of its circumcircle, then PA = PB + PC. Radhika's cyclic quadrilateral has sides 5, 5, 5 and 5 (a square); Brahmagupta gives √5 × 5 × 5 × 5 = 25, the same as 5 × 5.

A tangent touches the circle at exactly one point, and at that point one fact holds that you will use in every tangent question: the tangent is perpendicular to the radius at the point of contact. The radius is the shortest distance from the centre to the tangent line, and the shortest distance is always the perpendicular.

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How many tangents can you draw? From a point inside the circle, none. From a point on the circle, exactly one. From a point outside, exactly two. A circle as a whole has infinitely many tangents, and the tangents at the two ends of a diameter are parallel.

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Two tangents from an external point P touch the circle at A and B. Join OA, OB and OP:

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Special angles between the tangents. If ∠APB = 90°, OAPB is a square, OP = r√2 and PA = r. If ∠APB = 60°, then ∠APO = 30°, so OP = 2r and PA = r√3; also △APB is equilateral. If ∠APB = 120°, PA = r√3.

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Angles on the circle. If C is a point on the major arc, ∠ACB = 12∠AOB = 90° − 12∠APB. If C is on the minor arc, ∠ACB = 90° + 12∠APB.

Radhika stands 29 m from the centre of a circular pond of radius 20 m. The straight path from her that just touches the pond is √841 − 400 = 21 m long. That is the (20, 21, 29) triplet again.

This whole type stands on one line from Type 5: the two tangents from any external point are equal. When a circle sits inside a figure touching all its sides, every corner of the figure is an external point, so every corner sends out two equal tangent pieces.

Triangle with its incircle. Let the incircle of △ABC touch BC, CA, AB at D, E, F. Name the equal pieces: AE = AF = x, BF = BD = y, CD = CE = z. Then x + y = c, y + z = a, z + x = b, and x + y + z = s (the semi-perimeter). So

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One consequence saves a lot of work: a sum like PB + AR often collapses to a single side. If the incircle touches AB at P and AC at R, then AR = AP, so PB + AR = PB + PA = AB.

Right triangle. If ∠C = 90°, the two radii to the legs and the two tangent pieces from C make a square of side r. So both tangent pieces from C equal r, and writing the hypotenuse as (a − r) + (b − r) gives r = a + b − c2, where c is the hypotenuse. For the (8, 15, 17) triangle, r = 3; for (7, 24, 25), r = 3 as well.

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Quadrilateral around a circle. If a circle touches all four sides of ABCD, add the tangent pieces from the four corners: AB + CD = AD + BC, that is, the sums of opposite sides are equal. Hence a parallelogram that holds an incircle must be a rhombus, and in a trapezium with an incircle the two parallel sides together equal the two non-parallel sides together.

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Third tangent. Two tangents PA and PB are drawn from P. A third tangent touches the circle at Q (on the arc between A and B, facing P) and cuts PA at R and PB at S. Since RQ = RA and SQ = SB, the perimeter of △PRS = PR + RQ + QS + SP = PA + PB = 2PA, wherever Q lies on that arc.

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Excircle. A circle that touches BC and also AB and AC produced beyond B and C is the excircle opposite A. The tangent from A to it equals s, the semi-perimeter.

Radhika draws a triangle with sides 7, 8 and 9 and its incircle. s = 12, so the tangent pieces from the three corners are 12 − 7 = 5, 12 − 8 = 4 and 12 − 9 = 3. Check: 4 + 3 = 7, 5 + 3 = 8, 5 + 4 = 9.

Draw a tangent AT at A and a chord AB from the point of contact. The chord cuts the circle into two segments. The angle ∠BAT between the tangent and the chord sits in one segment; any angle ∠ACB made by the chord in the other segment is called the angle in the alternate segment.

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Why? Draw the diameter AD. ∠DAT = 90° (tangent ⊥ radius) and ∠ABD = 90° (angle in a semicircle). Then ∠BAT = 90° − ∠BAD = ∠ADB, and ∠ADB = ∠ACB because both stand on arc AB.

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The same idea in two quick forms:

Where it hides in a figure. (i) A tangent and a secant from an outside point: the angle between the tangent and a chord from the point of contact equals an angle of the inscribed triangle. (ii) A chord parallel to the tangent: alternate angles (parallel lines) and the alternate segment angle together make an isosceles triangle. (iii) A common tangent to two intersecting circles: if AB is the common tangent and the circles meet at C and D (C nearer to AB), then ∠ACB + ∠ADB = 180°.

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Radhika draws a tangent at A and a chord AB making 35° with it. Any point C on the far arc sees AB at 35°, and the centre sees it at 70°.

Take any point P and draw any line through it that cuts the circle at A and B. The product PA × PB comes out the same for every such line through P. This fixed product is called the power of P, and in the exam it gives three rules.

1. Chords crossing inside. Chords AB and CD meet at P inside the circle:

Proof: △PAC ~ △PDB, because ∠PAC = ∠PDB (same segment, arc CB) and the angles at P are vertically opposite. So PAPD = PCPB.

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2. Secants meeting outside. Lines through an outside point P cut the circle at A, B and at C, D:

Each length is measured from P, so it is the whole secant times its outer part.

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3. Tangent and secant. PT is a tangent and PAB a secant:

Think of the tangent as a secant whose two cutting points have come together.

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Radhika stands outside a circular fountain on a line that passes through it. The line enters the fountain 3 m from her and leaves it 27 m from her, so the tangent from her to the fountain is √3 × 27 = 9 m.

The common chord bisects the common tangent. If two circles cut at A and B, and the line AB meets a common tangent PQ at X, then XP² = XA × XB = XQ². So X is the midpoint of PQ.

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Always write the lengths from P. If the question gives a chord length (like AB = 5) and an outer part, convert it first: PB = PA + AB.

Now two circles come in together. Let their radii be r₁ ≥ r₂ and the distance between their centres be d. A line that touches both circles is a common tangent, and there are two kinds:

• Direct (external) common tangent: both circles lie on the same side of it, and it does not cross the line of centres between the circles.
• Transverse (internal) common tangent: the circles lie on opposite sides of it, and it crosses the line of centres between them.

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The number of common tangents depends only on how far apart the circles are. Compare d with r₁ + r₂ and r₁ − r₂:

Read the table like a story. Two far-apart circles have four tangents. Bring them closer until they touch, and the two transverse tangents merge into one, the tangent at the point of contact: three. Push them into each other and no line can pass between them, so only the two direct ones are left: two. When the small circle touches the big one from inside, only the tangent at that point remains: one. Push it fully inside and there are none. Concentric circles also have no common tangent.

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Three circles. For three separate, non-touching circles, each pair can have 4 common tangents, so the maximum total is 3 × 4 = 12.

Radhika takes circles of radii 7 cm and 3 cm. With the centres 12 cm apart (more than 10) there are 4 common tangents; at exactly 10 cm, 3; at 6 cm, 2; at exactly 4 cm, 1; at 2 cm, none.

Length of the direct common tangent (L). Let AB be a direct common tangent touching the circles (centres O₁, O₂) at A and B. O₁A ⊥ AB and O₂B ⊥ AB. Draw O₂K parallel to AB, meeting O₁A at K. Now ABO₂K is a rectangle, so O₂K = AB = L and O₁K = r₁ − r₂. Triangle O₁KO₂ is right-angled at K with hypotenuse d:

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Length of the transverse common tangent (M). Use the same construction. This time the radii point to opposite sides, so the short leg becomes r₁ + r₂:

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Subtracting and adding the squares gives two handy results: L² − M² = 4r₁r₂, and L² + M² = 2d² − 2(r₁² + r₂²). The direct tangent is always longer than the transverse one.

Circles touching externally. Here d = r₁ + r₂, so L = 2√r₁r₂. Also, if the circles touch at P and AB is a direct common tangent, the common tangent at P meets AB at T with TA = TP = TB (three equal tangents). So P lies on the circle with diameter AB, and ∠APB = 90°.

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Where the tangents cut the line of centres. The two direct common tangents meet the line O₁O₂ (produced) at a point dividing it externally in the ratio r₁ : r₂; the transverse tangents meet it at a point dividing O₁O₂ internally in the ratio r₁ : r₂. Both come from the similar triangles formed by the radii.

Common chord of two intersecting circles. The line of centres is the perpendicular bisector of the common chord. If the half chord is h, then d = √r₁² − h² + √r₂² − h² when the centres are on opposite sides of the chord, which is the usual case. If r₁, r₂ and d are given, the half chord is the height, drawn to the side d, of the triangle with sides r₁, r₂ and d: h = 2 × aread. Two equal circles of radius r passing through each other's centres have a common chord of r√3.

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Radhika's circles have radii 16 and 9 with centres 25 apart. They touch externally (16 + 9 = 25), so the direct tangent is 2√144 = 24, and the only transverse tangent is the one at the point of contact.

A third circle in the gap. Let two circles of radii r₁ and r₃ touch each other externally and also touch the same line. A smaller circle of radius r₂ that sits in the gap and touches both circles and the line satisfies 1√r₂ = 1√r₁ + 1√r₃. This follows from the tangent length 2√r₁r₃ between the two big circles being the sum of 2√r₁r₂ and 2√r₂r₃. With r₁ = 9 and r₃ = 4, the tangent length is 12 and 1√r₂ = 13 + 12 = 56, so r₂ = 1.44 cm.

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A quadrilateral has four sides and four angles. A diagonal cuts it into two triangles, so the angles of any quadrilateral add up to 360°. Most angle questions on a general quadrilateral end with this one line.

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Parallelogram (both pairs of opposite sides parallel):

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Rectangle = a parallelogram with all angles 90°. Its diagonals are equal and bisect each other.

Rhombus = a parallelogram with all sides equal.

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Square = rhombus + rectangle: equal sides, all angles 90°, equal diagonals meeting at 90°, and diagonal = a√2.

Trapezium (one pair of parallel sides, say AB ∥ CD, with diagonals meeting at O): the two angles on each slanting side are supplementary. The diagonals cut each other in the ratio of the parallel sides, and triangles AOB and COD are similar, so their areas are in the ratio (AB : CD)². An isosceles trapezium has equal slanting sides, equal base angles and equal diagonals, and it is cyclic.

Kite (two pairs of equal adjacent sides, AB = AD and CB = CD): the diagonal AC, joining the vertices where the equal sides meet, is the axis of symmetry. It bisects the other diagonal at right angles and bisects the angles at A and C; the angles at B and D are equal.

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Radhika's rhombus has diagonals 16 cm and 30 cm. The half-diagonals are 8 and 15, so each side is 17 cm (the 8-15-17 triplet), and the area is 12 × 16 × 30 = 240 cm².

Lengths inside a trapezium. Let the parallel sides be a (longer) and b (shorter). Join the midpoints E and F of the two slanting sides. This mid-segment EF is parallel to both bases and EF = a + b2, the average of the parallel sides. Draw the diagonal AC and you can see why: in △ACD the midpoint theorem gives b2, and in △ABC it gives a2.

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Radhika's trapezium has parallel sides 9 cm and 3 cm. Its mid-segment is 6 cm, the segment joining the midpoints of the diagonals is 3 cm, and the line through O parallel to the bases is 2 × 9 × 312 = 4.5 cm.

More diagonal results.

• Perpendicular diagonals (rhombus, kite, square, or any such quadrilateral): AB² + CD² = BC² + DA², because each side squared is the sum of the squares of two half-diagonal pieces. The area is 12d₁d₂. For any quadrilateral, area = 12d₁d₂ sin θ, where θ is the angle between the diagonals. If d₁ = 10, d₂ = 8 and θ = 30°, the area is 20.
• Mid-point quadrilateral. Joining the midpoints of the sides of any quadrilateral in order gives a parallelogram, each of whose sides is parallel to a diagonal and half its length. Its perimeter is d₁ + d₂, and its area is half that of the original. It is a rectangle if the diagonals are perpendicular, and a rhombus if they are equal. For diagonals 10 and 6, the perimeter is 16.
• Trapezium areas. With AB ∥ CD and diagonals meeting at O, △ABD and △ABC share the base AB and the same height, so ar(△AOD) = ar(△BOC). If AB : CD = 2 : 1, the four triangles AOB, AOD, BOC and COD have areas in the ratio 4 : 2 : 2 : 1.

A polygon with n sides can be cut into (n − 2) triangles by drawing the diagonals from one vertex. Each triangle gives 180°, so

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At every vertex the interior angle and the exterior angle make a straight line (180°). Walk once around any convex polygon and you turn through one full turn, so

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Regular polygon (all sides and all angles equal):

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Diagonals and triangles.

• A regular hexagon is made of 6 equilateral triangles; its side equals the radius of its circumcircle.

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Each interior angle of Radhika's regular polygon is 156°. The exterior angle is 24°, so n = 36024 = 15 sides, and it has 15 × 122 = 90 diagonals.

Regular hexagon and octagon. For a regular hexagon of side a, the area is 6 × √34a² = 3√32a². Its 3 long diagonals pass through the centre and each equals 2a, while its 6 short diagonals each equal a√3; together with the 6 sides these make the 9 diagonals above. For a = 4 cm: area = 24√3 ≈ 41.57 cm², long diagonal 8 cm, short diagonal 4√3 ≈ 6.93 cm. A regular octagon of side a has area 2(1 + √2)a². It can be cut from a square of side s by trimming four right-angled isosceles corners of equal legs x, where a = x√2 and s = x(2 + √2). For s = 10 cm, x ≈ 2.93 cm, a ≈ 4.14 cm, and the octagon's area is 100 − 4 × 12x² ≈ 82.84 cm².