Plane geometry — SSC CGL Quant Notes

Concepts and rules

The angles of a triangle add up to 180°. An exterior angle equals the sum of the two interior opposite angles. The larger side is always opposite the larger angle.

Centroid G is where the 3 medians meet. It divides every median in the ratio 2:1, the longer part being on the vertex side, so AG = 2/3 of AD and GD = 1/3 of AD.

The 3 medians cut a triangle into 6 small triangles of equal area. Hence area(GBC) = area(GCA) = area(GAB) = 1/3 of area(ABC).

Apollonius theorem: if AD is the median to BC, then AB² + AC² = 2(AD² + BD²). Equivalently 4AD² = 2AB² + 2AC² − BC². The sum of the squares of the medians is 3/4 of the sum of the squares of the sides.

Incentre I is where the internal angle bisectors meet; it is the centre of the incircle. ∠BIC = 90° + A/2 always. Inradius r = Area ÷ s, where s is the semi-perimeter. For a right triangle, r = (sum of legs − hypotenuse)/2.

Angle bisector theorem: if the bisector of ∠A meets BC at D, then BD : DC = AB : AC. The external bisector divides BC externally in the same ratio. The excentre opposite A gives ∠BI₁C = 90° − A/2.

Circumcentre O is where the perpendicular bisectors of the sides meet; OA = OB = OC = R. R = abc ÷ (4 × Area). If ∠A is acute, ∠BOC = 2A; if ∠A is obtuse, ∠BOC = 360° − 2A. In a right triangle O is the midpoint of the hypotenuse, so R = hypotenuse/2.

Orthocentre H is where the altitudes meet. If ∠B and ∠C are both acute, ∠BHC = 180° − A (this covers acute A and obtuse A). If ∠B or ∠C is obtuse, ∠BHC = A. In a right triangle H is the right-angle vertex.

Equilateral triangle of side a: all 4 centres coincide; height = √3a/2, area = √3a²/4, R = a/√3, r = a/(2√3), so R = 2r. In any triangle O, G and H lie on the Euler line with HG : GO = 2:1.

Similarity tests: AA (2 equal angles), SSS (all sides in the same ratio), SAS (2 sides in the same ratio with equal included angle). Write the vertices in matching order, e.g. ΔABC ~ ΔPQR means A↔P, B↔Q, C↔R.

If the side ratio of 2 similar triangles is k, then perimeters, medians, altitudes, angle bisectors and inradii are all in ratio k, but areas are in ratio k².

Basic proportionality (Thales): if DE ∥ BC with D on AB and E on AC, then AD/DB = AE/EC and ΔADE ~ ΔABC with ratio AD/AB. Midpoint theorem: the segment joining midpoints of 2 sides is parallel to the third side and half of it.

Right triangle with ∠A = 90° and altitude AD to BC: AD² = BD × DC, AB² = BD × BC, AC² = CD × CB, and AD = AB × AC ÷ BC.

Method and shortcuts

Step 1: Name the centre in the question (incentre, circumcentre, orthocentre, centroid) and write its key fact before touching numbers.

Step 2: For angle questions use the centre's formula at BC. Check whether ∠A is acute or obtuse for the circumcentre, and whether ∠B or ∠C is obtuse for the orthocentre.

Step 3: For length or area questions with parallel lines, find the pair of similar triangles, write the side ratio k in matching order, then square it for areas.

Shortcut: area of a trapezium cut off by DE ∥ BC = area(ABC) × (1 − k²), where k = AD/AB. It works only when DE is parallel to BC.

Shortcut: the median to the hypotenuse of a right triangle equals half the hypotenuse, because the circumcentre is the midpoint of the hypotenuse. It fails for non-right triangles; use Apollonius there.

Why ∠BIC = 90° + A/2: in ΔIBC the angles at B and C are B/2 and C/2, so ∠BIC = 180° − (B + C)/2 = 180° − (180° − A)/2.

Example 1: angle at the incentre

Question: In ΔABC, ∠A = 70° and I is the incentre. Find ∠BIC.

Solution: ∠BIC = 90° + A/2 = 90° + 35° = 125°.

Answer: 125°

Example 2: circumcentre and orthocentre

Question: In an acute-angled ΔABC, ∠A = 64°. O is the circumcentre and H is the orthocentre. Find ∠BOC and ∠BHC.

Solution: ∠A is acute, so ∠BOC = 2 × 64° = 128°. All angles are acute, so ∠BHC = 180° − 64° = 116°.

Answer: ∠BOC = 128°, ∠BHC = 116°

Example 3: angle bisector theorem

Question: In ΔABC, AB = 12 cm, AC = 8 cm and BC = 15 cm. The bisector of ∠A meets BC at D. Find BD and DC.

Solution: BD : DC = AB : AC = 12 : 8 = 3 : 2. So BD = 15 × 3/5 = 9 cm and DC = 15 × 2/5 = 6 cm.

Answer: BD = 9 cm, DC = 6 cm

Example 4: parallel line and area

Question: In ΔABC, D and E lie on AB and AC with DE ∥ BC. AD = 4 cm, DB = 6 cm and area(ΔADE) = 32 cm². Find the area of trapezium DBCE. (a) 200 cm² (b) 136 cm² (c) 168 cm² (d) 150 cm²

Solution: ΔADE ~ ΔABC with k = AD/AB = 4/10 = 2/5. Area ratio = 4/25, so area(ABC) = 32 × 25/4 = 200 cm².

Trapezium = 200 − 32 = 168 cm². Option (a) is the trap of stopping at the whole triangle.

Answer: (c) 168 cm²

Example 5: median and centroid by Apollonius

Question: In ΔABC, AB = 7 cm, BC = 8 cm and CA = 9 cm. G is the centroid. Find AG. (a) 7 cm (b) 14/3 cm (c) 7/3 cm (d) 16/3 cm

Solution: Let AD be the median to BC. 4AD² = 2AB² + 2AC² − BC² = 2 × 49 + 2 × 81 − 64 = 98 + 162 − 64 = 196, so AD² = 49 and AD = 7 cm.

G divides AD in 2:1 from A, so AG = 2/3 × 7 = 14/3 cm. Option (a) is the whole median and (c) is GD.

Answer: (b) 14/3 cm

Example 6: altitude on hypotenuse, inradius and circumradius

Question: In ΔABC, ∠A = 90° and AD ⊥ BC. If BD = 9 cm and DC = 16 cm, find r + R, where r is the inradius and R the circumradius of ΔABC. (a) 15 cm (b) 16.5 cm (c) 18 cm (d) 17.5 cm

Solution: BC = 9 + 16 = 25. AB² = BD × BC = 9 × 25 = 225, so AB = 15. AC² = CD × CB = 16 × 25 = 400, so AC = 20. (Also AD² = 9 × 16 = 144, AD = 12.)

r = (15 + 20 − 25)/2 = 5 cm. R = hypotenuse/2 = 25/2 = 12.5 cm. r + R = 17.5 cm.

Answer: (d) 17.5 cm

Common mistakes

Using the side ratio as the area ratio. Areas go as the square: side ratio 2:3 means area ratio 4:9.

Reading the centroid ratio from the wrong end. AG : GD = 2:1, so the part next to the side is the shorter one.

Writing ∠BOC = 2A when ∠A is obtuse. Then O lies on the other side of BC and ∠BOC = 360° − 2A.

Mixing the incentre and excentre formulas: incentre gives 90° + A/2, excentre gives 90° − A/2, orthocentre gives 180° − A.

In the proportionality theorem, using AD/AB = AE/EC. Keep like with like: AD/DB = AE/EC, or AD/AB = AE/AC = DE/BC.

Assuming the angle bisector cuts BC in half. That happens only when AB = AC.

Practice questions

Q1. G is the centroid of ΔABC, whose area is 96 cm². Find the area of ΔAGB. (a) 32 cm² (b) 24 cm² (c) 48 cm² (d) 16 cm²

Q2. In an acute-angled ΔABC, I is the incentre and O the circumcentre. If ∠BIC = 118°, find ∠BOC. (a) 118° (b) 124° (c) 104° (d) 112°

Q3. ΔABC ~ ΔPQR and their areas are 144 cm² and 81 cm². If an altitude of ΔABC is 8 cm, the corresponding altitude of ΔPQR is: (a) 4.5 cm (b) 6 cm (c) 5.4 cm (d) 6.4 cm

Answers

Q1. (a) The 3 triangles GBC, GCA, GAB are equal, so area(AGB) = 96/3 = 32 cm².

Q2. (d) 90° + A/2 = 118° gives A/2 = 28°, A = 56°. ∠BOC = 2 × 56° = 112°.

Q3. (b) Side ratio = √(144/81) = 12/9 = 4/3. Altitude of ΔPQR = 8 × 3/4 = 6 cm.

Quick revision

∠BIC = 90° + A/2; ∠BOC = 2A (A acute); ∠BHC = 180° − A (B, C acute); excentre ∠BI₁C = 90° − A/2.

Centroid: 2:1 from the vertex; 6 equal small triangles; 4AD² = 2b² + 2c² − a².

r = Area/s, R = abc/(4 × Area); right triangle: R = hypotenuse/2, r = (a + b − c)/2.

Similar triangles: lengths in ratio k, areas in ratio k². Bisector: BD : DC = AB : AC.

Concepts and rules

The perpendicular from the centre to a chord bisects the chord. If a chord of length 2L is at distance d from the centre of a circle of radius r, then L² + d² = r².

Equal chords are equidistant from the centre and subtend equal angles at the centre. A longer chord is nearer to the centre; the diameter is the longest chord.

The angle an arc subtends at the centre is twice the angle it subtends at any point on the remaining part of the circle. Angles in the same segment are equal. The angle in a semicircle is 90°.

If chord AB subtends ∠AOB = θ at the centre, a point on the major arc sees AB at θ/2 and a point on the minor arc sees it at 180° − θ/2.

Cyclic quadrilateral: opposite angles add to 180°, and an exterior angle equals the interior opposite angle. A parallelogram that is cyclic must be a rectangle.

A tangent is perpendicular to the radius at the point of contact. From an external point P at distance d from the centre, tangent length PT = √(d² − r²). The 2 tangents from P are equal.

If tangents PA and PB touch the circle at A and B, then ∠APB + ∠AOB = 180°, OP bisects ∠APB, and quadrilateral OAPB has area 2 × (1/2) × r × PA = r × PA.

Alternate segment theorem: the angle between a tangent and a chord through the point of contact equals the angle that chord subtends in the alternate segment.

Intersecting chords: if chords AB and CD meet at P inside the circle, PA × PB = PC × PD. If 2 secants from an outside point P meet the circle at A, B and C, D, then PA × PB = PC × PD, and for a tangent PT, PT² = PA × PB.

A quadrilateral circumscribing a circle (all 4 sides tangent) satisfies AB + CD = AD + BC, because the tangent lengths from each vertex are equal.

Method and shortcuts

Step 1: Join the centre to the points that matter: to the ends of a chord, to the point of contact of a tangent. Most circle problems become right triangles after this.

Step 2: Mark every right angle (radius ⊥ tangent, angle in a semicircle, centre ⊥ chord) and every pair of equal lengths (radii, tangents from one point).

Step 3: For angles, find the arc each angle stands on; angles on the same arc are equal and the centre angle is double.

Shortcut: 2 parallel chords on opposite sides of the centre are d₁ + d₂ apart; on the same side they are |d₁ − d₂| apart. If the question does not say which, both answers may appear in the options.

Shortcut: angle between 2 tangents = 180° − angle at the centre between the radii to the contact points.

Example 1: distance of a chord

Question: A chord of length 16 cm is drawn in a circle of radius 10 cm. Find its distance from the centre.

Solution: Half chord = 8. d² = 10² − 8² = 100 − 64 = 36, so d = 6 cm.

Answer: 6 cm

Example 2: two parallel chords

Question: In a circle of radius 13 cm, 2 parallel chords of lengths 10 cm and 24 cm lie on opposite sides of the centre. Find the distance between them.

Solution: For 10 cm: d₁ = √(169 − 25) = 12. For 24 cm: d₂ = √(169 − 144) = 5. Opposite sides, so distance = 12 + 5 = 17 cm (on the same side it would be 7 cm).

Answer: 17 cm

Example 3: centre angle and both arcs

Question: O is the centre and ∠AOB = 130°. C is on the major arc and D on the minor arc. Find ∠ACB and ∠ADB.

Solution: ∠ACB = 130°/2 = 65°. ACBD is cyclic, so ∠ADB = 180° − 65° = 115°.

Answer: ∠ACB = 65°, ∠ADB = 115°

Example 4: tangents and quadrilateral area

Question: From a point P, 17 cm from the centre O of a circle of radius 8 cm, tangents PA and PB are drawn. Find the area of quadrilateral OAPB. (a) 60 cm² (b) 136 cm² (c) 240 cm² (d) 120 cm²

Solution: PA = √(17² − 8²) = √(289 − 64) = √225 = 15 cm. ΔOAP is right-angled at A, area = (1/2) × 8 × 15 = 60.

OAPB is 2 such triangles: 2 × 60 = 120 cm². Option (a) counts only 1 triangle.

Answer: (d) 120 cm²

Example 5: tangent–secant and chord distance

Question: From an external point P, a tangent PT = 12 cm is drawn to a circle of radius 13 cm, and a secant PAB cuts the circle at A and B with PA = 8 cm. Find the distance of chord AB from the centre. (a) 12 cm (b) 5 cm (c) 10 cm (d) 13 cm

Solution: PT² = PA × PB gives 144 = 8 × PB, so PB = 18 and AB = 18 − 8 = 10 cm.

Half chord = 5, so distance = √(13² − 5²) = √(169 − 25) = √144 = 12 cm. Option (c) is the chord length itself.

Answer: (a) 12 cm

Example 6: alternate segment with 2 tangents

Question: PA and PB are tangents from P to a circle with centre O. C is a point on the major arc AB with ∠ACB = 58°. Find ∠APB. (a) 58° (b) 122° (c) 64° (d) 32°

Solution: Centre angle ∠AOB = 2 × 58° = 116°. In quadrilateral OAPB the angles at A and B are 90° each, so ∠APB = 360° − 90° − 90° − 116° = 64°.

Check by the alternate segment theorem: ∠PAB = ∠ACB = 58° and PA = PB, so ∠PBA = 58° and ∠APB = 180° − 116° = 64°.

Answer: (c) 64°

Common mistakes

Using the full chord instead of half the chord in L² + d² = r². The right triangle is formed with half the chord.

Taking the inscribed angle as θ/2 when the point lies on the minor arc. There it is 180° − θ/2.

Answering only d₁ + d₂ for parallel chords when the chords are on the same side of the centre; then the distance is |d₁ − d₂|.

In PT² = PA × PB, using AB in place of PB. Both lengths are measured from P: the near point A and the far point B.

Forgetting that the radius meets the tangent at 90°; without this right angle you cannot use Pythagoras.

Practice questions

Q1. Chords AB and CD of a circle intersect at P inside the circle. If PA = 6 cm, PB = 4 cm and PC = 3 cm, find PD. (a) 6 cm (b) 8 cm (c) 4.5 cm (d) 2 cm

Q2. A quadrilateral ABCD circumscribes a circle. AB = 7 cm, BC = 9 cm and CD = 8 cm. Find AD. (a) 8 cm (b) 7 cm (c) 5 cm (d) 6 cm

Q3. A chord of length 30 cm is at a distance of 8 cm from the centre of a circle. Find the length of a chord at a distance of 15 cm from the centre. (a) 16 cm (b) 18 cm (c) 12 cm (d) 20 cm

Answers

Q1. (b) PA × PB = PC × PD gives 6 × 4 = 3 × PD, so PD = 8 cm.

Q2. (d) AB + CD = AD + BC gives 7 + 8 = AD + 9, so AD = 6 cm.

Q3. (a) r² = 15² + 8² = 289, r = 17. Half chord = √(289 − 225) = 8, chord = 16 cm.

Quick revision

Half-chord² + distance² = r²; tangent² = d² − r²; tangent ⊥ radius.

Centre angle = 2 × angle on the major arc; minor-arc angle = 180° − half the centre angle; semicircle angle = 90°.

Cyclic quadrilateral: opposite angles sum to 180°. Circumscribed quadrilateral: AB + CD = AD + BC.

PA × PB = PC × PD for chords and secants; PT² = PA × PB for a tangent and a secant.

Concepts and rules

2 triangles are congruent when one fits exactly on the other: all 3 sides and all 3 angles match. Write the vertices in matching order: ΔABC ≅ ΔPQR means AB = PQ, BC = QR, CA = RP and ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R.

SSS: 3 sides of one equal to 3 sides of the other.

SAS: 2 sides and the angle included between them are equal. The angle must lie between the 2 sides.

ASA: 2 angles and the side between them are equal. AAS: 2 angles and a non-included side are equal (it reduces to ASA, since the third angle is then equal too).

RHS: in right triangles, the hypotenuse and 1 other side are equal. This is the only case where 2 sides and a non-included angle are enough.

AAA is not a congruence test: equal angles give only similar triangles of possibly different size. SSA (2 sides and a non-included angle) is not a test either, because 2 different triangles can fit the data.

CPCT: once congruence is proved, every pair of corresponding parts is equal, including medians, altitudes and angle bisectors drawn to corresponding sides.

Isosceles triangle (AB = AC): ∠B = ∠C, and the converse is true. The median, altitude and angle bisector from A are the same line, and it is the perpendicular bisector of BC.

Every point on the perpendicular bisector of a segment is equidistant from its ends. Every point on the bisector of an angle is equidistant (by perpendicular distance) from its arms. Both are proved by congruence.

Triangle inequality: each side is less than the sum and more than the difference of the other 2. With sides a and b, the third side c satisfies |a − b| < c < a + b.

Triangles with the same 3 side lengths are congruent (SSS), so counting non-congruent triangles means counting different side triples a ≤ b ≤ c with a + b > c.

Method and shortcuts

Step 1: Mark what is given equal and what is common (a shared side, vertically opposite angles, angles between parallels).

Step 2: Match the 3 facts to SSS, SAS, ASA, AAS or RHS. If the angle is not between the 2 sides and the triangles are not right-angled, no test applies.

Step 3: Use CPCT to transfer a length or angle, then finish with Pythagoras, angle sum or the isosceles property.

Shortcut: from a point on an angle bisector, the 2 perpendiculars to the arms are equal, and the feet are equally far from the vertex (AAS). This turns many bisector problems into simple subtraction.

Shortcut for counting integer third sides: the count is (a + b − 1) − (|a − b| + 1) + 1 = 2 × min(a, b) − 1.

Example 1: naming the test

Question: In ΔABC and ΔPQR, AB = PQ, BC = QR and ∠B = ∠Q. Which test proves congruence?

Solution: ∠B lies between sides AB and BC, and ∠Q between PQ and QR. 2 sides with the included angle: SAS.

Answer: SAS

Example 2: isosceles angles

Question: In ΔABC, AB = AC and ∠A = 40°. Find ∠B.

Solution: ∠B = ∠C, so 2∠B = 180° − 40° = 140° and ∠B = 70°.

Answer: 70°

Example 3: possible third sides

Question: 2 sides of a triangle are 5 cm and 9 cm. How many integer values can the third side take?

Solution: 9 − 5 < c < 9 + 5, so 4 < c < 14 and c = 5, 6, …, 13. That is 13 − 5 + 1 = 9 values (shortcut 2 × 5 − 1 = 9).

Answer: 9

Example 4: congruence inside an isosceles triangle

Question: In ΔABC, AB = AC = 17 cm and BC = 16 cm. D is the midpoint of BC. Find the area of ΔABC. (a) 136 cm² (b) 120 cm² (c) 240 cm² (d) 128 cm²

Solution: ΔABD ≅ ΔACD by SSS (AB = AC, BD = DC, AD common), so ∠ADB = ∠ADC; together they make 180°, so each is 90°.

AD = √(17² − 8²) = √(289 − 64) = 15. Area = (1/2) × 16 × 15 = 120 cm².

Answer: (b) 120 cm²

Example 5: angle bisector and equal perpendiculars

Question: In ΔABC, ∠B = 90°, AB = 6 cm and BC = 8 cm. The bisector of ∠A meets BC at D, and DE ⊥ AC. Find DE. (a) 4 cm (b) 2.4 cm (c) 3 cm (d) 3.6 cm

Solution: AC = √(36 + 64) = 10. ΔABD ≅ ΔAED by AAS (equal angles at A, right angles at B and E, AD common), so DB = DE = x and AE = AB = 6, giving EC = 4.

In right ΔDEC: DC = 8 − x, so (8 − x)² = x² + 4², i.e. 64 − 16x = 16, x = 3 cm. Check by the bisector theorem: BD : DC = 6 : 10, BD = 8 × 6/16 = 3.

Answer: (c) 3 cm

Example 6: counting non-congruent triangles

Question: How many non-congruent triangles with integer sides have perimeter 15 cm? (a) 5 (b) 6 (c) 8 (d) 7

Solution: Take a ≤ b ≤ c with a + b + c = 15. a + b > c means c < 7.5, so c ≤ 7; and c ≥ 5 since c is the largest.

c = 7: (1, 7, 7), (2, 6, 7), (3, 5, 7), (4, 4, 7). c = 6: (3, 6, 6), (4, 5, 6). c = 5: (5, 5, 5). Total 4 + 2 + 1 = 7.

Answer: (d) 7

Common mistakes

Accepting AAA as congruence. Triangles with angles 50°, 60°, 70° can be of any size; AAA gives similarity only.

Calling 2 sides and any angle SAS. If the angle is not between the 2 sides it is SSA, which proves nothing unless the triangles are right-angled (RHS).

Matching vertices in the wrong order and then equating the wrong sides by CPCT. Write ΔABC ≅ ΔPQR only after checking which vertex goes to which.

Using ≤ in the triangle inequality. A side equal to the sum of the other 2 gives a straight line, not a triangle.

Counting (2, 6, 7) and (6, 2, 7) as different triangles. Order does not matter; list sides with a ≤ b ≤ c.

Practice questions

Q1. Which of the following is NOT a test of congruence of triangles? (a) SSS (b) SAS (c) AAA (d) RHS

Q2. In ΔABC, AB = AC and ∠B = 2∠A. Find ∠B. (a) 36° (b) 54° (c) 72° (d) 108°

Q3. 2 sides of a triangle are 4 cm and 10 cm. How many integer lengths are possible for the third side? (a) 7 (b) 8 (c) 6 (d) 9

Answers

Q1. (c) Equal angles fix only the shape, not the size; a triangle and its enlargement have the same angles.

Q2. (c) Let ∠A = x. Then ∠B = ∠C = 2x, so x + 2x + 2x = 180°, x = 36° and ∠B = 72°.

Q3. (a) 6 < c < 14 gives c = 7 to 13, which is 7 values (2 × 4 − 1 = 7).

Quick revision

Valid tests: SSS, SAS (included angle), ASA, AAS, RHS. Invalid: AAA, SSA.

Isosceles: equal sides ⇔ equal base angles; the apex median is also the altitude and bisector.

A point on an angle bisector is equidistant from both arms; a point on a perpendicular bisector is equidistant from both ends.

Third side: |a − b| < c < a + b; integer count = 2 × min(a, b) − 1.

Concepts and rules

Let 2 circles have radii r₁ ≥ r₂ and let d be the distance between their centres. A common tangent touches both circles.

Direct (external) common tangent: both circles lie on the same side of it. Length between the 2 points of contact = √(d² − (r₁ − r₂)²).

Transverse (internal) common tangent: it passes between the circles. Length between the points of contact = √(d² − (r₁ + r₂)²). It exists only when d > r₁ + r₂.

Why: shift the tangent parallel to itself through the centre of the smaller circle. You get a right triangle with hypotenuse d and one leg r₁ − r₂ (direct) or r₁ + r₂ (transverse); the other leg is the tangent length.

Number of common tangents: d > r₁ + r₂ (separate) gives 4; d = r₁ + r₂ (touch externally) gives 3; r₁ − r₂ < d < r₁ + r₂ (intersect at 2 points) gives 2; d = r₁ − r₂ with d > 0 (touch internally) gives 1; d < r₁ − r₂ (one inside the other, including concentric) gives 0.

If the circles touch externally, d = r₁ + r₂ and the direct common tangent = √((r₁ + r₂)² − (r₁ − r₂)²) = 2√(r₁r₂).

The 2 direct common tangents (for unequal circles) meet on the line of centres at a point P that divides the centre line externally in the ratio r₁ : r₂, i.e. PO₁ : PO₂ = r₁ : r₂. The 2 transverse tangents meet at a point dividing O₁O₂ internally in r₁ : r₂.

Equal circles have parallel direct common tangents, each of length d, so the direct tangents never meet.

3 circles touching each other externally with radii a, b, c have centres forming a triangle with sides a + b, b + c, c + a.

If 2 circles of radii a and b touch each other externally and both touch a straight line, a third circle touching both and the same line in the gap between them has radius r with 1/√r = 1/√a + 1/√b.

Method and shortcuts

Step 1: Write r₁, r₂ and d. Compare d with r₁ + r₂ and r₁ − r₂ to know the position of the circles.

Step 2: Decide which tangent is asked: direct uses the difference of radii, transverse uses the sum. A memory aid: transverse crosses between the circles, so the radii add up.

Step 3: For reverse questions (find d or a radius), square the formula: d² = tangent² + (r₁ ∓ r₂)².

Shortcut: for circles that touch externally, the direct tangent is 2√(r₁r₂) and there is no transverse tangent segment of positive length (its length is 0).

Shortcut: for the meeting point P of direct tangents, PO₂ = d × r₂/(r₁ − r₂) beyond the smaller circle; for transverse tangents the point is at d × r₁/(r₁ + r₂) from O₁.

Example 1: direct common tangent

Question: 2 circles of radii 8 cm and 3 cm have centres 13 cm apart. Find the length of a direct common tangent.

Solution: √(13² − (8 − 3)²) = √(169 − 25) = √144 = 12 cm.

Answer: 12 cm

Example 2: transverse common tangent

Question: 2 circles of radii 10 cm and 5 cm have centres 17 cm apart. Find the length of a transverse common tangent.

Solution: 17 > 10 + 5, so it exists. √(17² − 15²) = √(289 − 225) = √64 = 8 cm.

Answer: 8 cm

Example 3: number of common tangents

Question: Circles of radii 7 cm and 4 cm have centres 11 cm apart. How many common tangents can be drawn?

Solution: d = 11 = 7 + 4, so the circles touch externally: 2 direct tangents and 1 tangent at the point of contact, total 3.

Answer: 3

Example 4: touching circles

Question: 2 circles of radii 9 cm and 4 cm touch each other externally. Find the length of their direct common tangent. (a) 13 cm (b) 12 cm (c) 6 cm (d) 10 cm

Solution: d = 9 + 4 = 13. Length = √(13² − 5²) = √144 = 12 cm, the same as 2√(9 × 4) = 2 × 6 = 12 cm. Option (a) is d itself and (c) forgets the factor 2.

Answer: (b) 12 cm

Example 5: meeting point of direct tangents

Question: Circles with centres O₁ and O₂ have radii 6 cm and 2 cm, and O₁O₂ = 12 cm. Their direct common tangents meet at P. Find the length of the tangent from P to the larger circle. (a) 12√2 cm (b) 18 cm (c) 6√2 cm (d) 16 cm

Solution: P lies beyond O₂ with PO₁ : PO₂ = 6 : 2 = 3 : 1. Let PO₂ = x; then PO₁ = x + 12 = 3x, so x = 6 and PO₁ = 18 cm.

Tangent to the larger circle = √(18² − 6²) = √(324 − 36) = √288 = 12√2 cm. Check: tangent to the smaller circle = √(36 − 4) = 4√2, and the difference 8√2 equals the direct tangent √(144 − 16) = √128.

Answer: (a) 12√2 cm

Example 6: circle in the gap

Question: 2 circles of radii 9 cm and 4 cm touch each other externally and both touch a straight line. A smaller circle touches both circles and the same line. Find its radius. (a) 1.5 cm (b) 1.2 cm (c) 1.44 cm (d) 2.25 cm

Solution: Contact points on the line are 2√(ab) apart for touching circles. For the 3 pairs: 2√(9 × 4) = 2√(9r) + 2√(4r), i.e. 6 = 3√r + 2√r = 5√r.

So √r = 6/5 and r = 36/25 = 1.44 cm (same as 1/√r = 1/3 + 1/2 = 5/6).

Answer: (c) 1.44 cm

Common mistakes

Swapping the formulas. Direct uses (r₁ − r₂), transverse uses (r₁ + r₂); the transverse tangent is always the shorter one.

Computing a transverse tangent when d ≤ r₁ + r₂. If the circles touch or intersect, there is no transverse tangent segment of positive length.

Counting 4 tangents for touching circles. External touch gives 3, internal touch gives 1.

Using d as the gap between the circles (edge to edge). d is always centre to centre; gap = d − r₁ − r₂.

In reverse problems, forgetting that r₁ − r₂ = k gives r₁ = r₂ + k, not r₂ − k, when r₂ is the smaller radius.

Practice questions

Q1. 2 circles of radii 11 cm and 5 cm have centres 10 cm apart. How many common tangents do they have? (a) 4 (b) 2 (c) 3 (d) 1

Q2. The transverse common tangent of 2 circles of radii 5 cm and 3 cm is 15 cm long. Find the distance between their centres. (a) 15 cm (b) 16 cm (c) 17 cm (d) 18 cm

Q3. The centres of 2 circles are 25 cm apart and their direct common tangent is 24 cm. The smaller circle has radius 3 cm. Find the radius of the larger circle. (a) 7 cm (b) 13 cm (c) 12 cm (d) 10 cm

Answers

Q1. (b) 11 − 5 = 6 < 10 < 16 = 11 + 5, so the circles intersect and have 2 common tangents.

Q2. (c) d² = 15² + (5 + 3)² = 225 + 64 = 289, d = 17 cm.

Q3. (d) (r₁ − 3)² = 25² − 24² = 49, so r₁ − 3 = 7 and r₁ = 10 cm.

Quick revision

Direct: √(d² − (r₁ − r₂)²). Transverse: √(d² − (r₁ + r₂)²).

Tangent count: separate 4, external touch 3, intersecting 2, internal touch 1, one inside the other 0.

Touching externally: direct tangent = 2√(r₁r₂).

Direct tangents meet where PO₁ : PO₂ = r₁ : r₂ (external division); circle in the gap: 1/√r = 1/√a + 1/√b.