Constructions (Circles)

Q: Why does the circle drawn on OP as diameter give the points of contact of the tangents from P?

Any point A on that circle makes ∠ OAP = 90° (angle in a semicircle). A line through a point of the given circle perpendicular to the radius there is a tangent, so the points common to both circles are the points of contact.

Q: A point P is 10 cm from the centre of a circle of radius 6 cm. Find the length of each tangent from P.

√(102 - 62) = √(64) = 8 cm.

Q: Two tangents PA and PB are drawn from P to a circle with centre O, and ∠ AOB = 110°. Find ∠ APB.

∠ APB = 180° - 110° = 70°, since ∠ OAP = ∠ OBP = 90° in quadrilateral OAPB.

ICSE Class 10 · Mathematics · Chapter 19

What this chapter covers

Instruments, accuracy and the basic constructions you need
Tangent at a point on a circle
Tangents to a circle from an external point
Circumscribing a circle on a triangle
Inscribing a circle in a triangle
The regular hexagon and its two circles
Circumscribing a circle on a regular hexagon
Inscribing a circle in a regular hexagon
Writing a construction answer in the examination

Key points

A tangent is perpendicular to the radius at the point of contact; the tangent at a point P is the perpendicular to OP at P.
Tangents from an external point P: bisect OP at M, draw the circle with centre M through O and P, and join P to the two points where it meets the given circle.
The construction works because the angle in a semicircle is 90^°; each tangent has length √(d² - r²), and the two tangents are equal.
The angle between two tangents and the angle they subtend at the centre add up to 180^°.
Circumcircle of a triangle: centre where the perpendicular bisectors of the sides meet; radius to a vertex; centre inside, on (midpoint of hypotenuse) or outside for acute, right and obtuse triangles.
Incircle of a triangle: centre where the angle bisectors meet; radius is the perpendicular to a side; centre always inside; check with r = Δ / s.
A regular hexagon has interior angles of 120^° and splits into six equilateral triangles from its centre.
For a regular hexagon of side a: circumradius R = a; inradius r = (√(3))/(2)a; the incircle touches each side at its midpoint.
Keep all construction arcs, draw a rough figure first, and state every measurement with its unit to one decimal place.

Questions answered

Define a tangent to a circle and its point of contact.

A tangent to a circle is a straight line that meets the circle at exactly one point; that point is the point of contact.

Why does the circle drawn on OP as diameter give the points of contact of the tangents from P?

Any point A on that circle makes ∠ OAP = 90^° (angle in a semicircle). A line through a point of the given circle perpendicular to the radius there is a tangent, so the points common to both circles are the points of contact.

A point P is 10 cm from the centre of a circle of radius 6 cm. Find the length of each tangent from P.

√(10² - 6²) = √(64) = 8 cm.

A point lies 2 cm from the centre of a circle of radius 3 cm. How many tangents can be drawn from it? Give a reason.

None, because the point lies inside the circle; every line through it cuts the circle at two points.

Two tangents PA and PB are drawn from P to a circle with centre O, and ∠ AOB = 110^°. Find ∠ APB.

∠ APB = 180^° - 110^° = 70^°, since ∠ OAP = ∠ OBP = 90^° in quadrilateral OAPB.

Which lines are constructed to locate the circumcentre of a triangle, and why?

The perpendicular bisectors of two sides, because every point on the perpendicular bisector of a side is equidistant from its ends, so their meeting point is equidistant from all three vertices.

Where does the circumcentre of a right-angled triangle lie? Find the circumradius when the legs are 6 cm and 8 cm.

At the midpoint of the hypotenuse. The hypotenuse is 10 cm, so the circumradius is 5 cm.

Why must a perpendicular be drawn from the incentre to a side before drawing the incircle?

The radius of the incircle is the shortest (perpendicular) distance from the incentre to a side; the distance to a vertex is longer, and a circle of that radius would cut the sides.

Past board exam questions

ICSE 2027 sample paper · 4 marks

Use ruler and compass for this question: (a) Draw line segment AB of length 6 cm. (b) Construct the locus of points which are equidistant from A and B. Mark a point on the locus which is 4.5 cm from A. Name this point as O. (c) Construct the locus of all points which are 4.5 cm from O. (d) Mark the point of intersection of loci (b) and (c) as P, which is on the same side of AB as O. (e) Construct the locus of points which are equidistant from AP and AB.

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Steps of construction: (a) Draw AB = 6 cm. (b) The locus of points equidistant from A and B is the perpendicular bisector of AB: with A and B as centres and radius more than 3 cm, draw arcs on both sides of AB meeting at two points; join them. With A as centre and radius 4.5 cm, cut this bisector at O (OA = OB = 4.5 cm). (c) The locus of points 4.5 cm from O is the circle with centre O and radius 4.5 cm (it passes through A and B). (d) The perpendicular bisector meets the circle at two points; the one on the same side of AB as O (farther from AB) is P. Join AP. (e) The locus of points equidistant from AP and AB is the bisector of ∠PAB: with A as centre draw an arc cutting AB and AP; …

ICSE 2026 sample paper · 5 marks

Use a ruler and compass to answer this question. (a) Construct a circle of radius 4.5cm and draw a chord AB of length 6.5 cm. (b) At A, construct ∠CAB=75°, where C lies on the circumference of the circle. (c) Construct the locus of all points equidistant from A and B. (d) Construct the locus of all points equidistant from CA and BA. (e) Mark the point of intersection of the two loci as P. Measure and write down the length of CP.

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Steps of construction: (a) Draw a circle with centre O and radius 4.5 cm. With A on the circle as centre and radius 6.5 cm, cut the circle at B; join AB. (b) At A construct ∠BAX = 75° (60° + half of the angle between 60° and 90°), on the side of AB containing the centre; let AX meet the circle at C. (c) The locus of points equidistant from A and B is the perpendicular bisector of AB (it passes through O). (d) The locus of points equidistant from CA and BA is the bisector of ∠CAB. (e) The perpendicular bisector of AB and the bisector of ∠CAB meet at P. On measuring, CP ≈ 5.1 cm (calculation gives 5.1 cm; the board's draft marking scheme gives CP = 4.9 cm, within construction tolerance).

ICSE 2026 board paper · 5 marks

Use ruler and compass for the following construction: (a) construct an equilateral triangle ABC of side 5 cm. (b) construct the circumcircle of ΔABC. (c) construct the locus of points which are equidistant from AB and BC. Mark the point where the circumcircle and locus meet, as D. (d) give the geometrical name of quadrilateral ABCD.

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(a) Draw BC = 5 cm. With B and C as centres and radius 5 cm, draw arcs meeting at A. Join AB and AC. (b) Draw the perpendicular bisectors of AB and BC; they meet at O, the circumcentre. With centre O and radius OA (= (5)/(√(3)) ≈ 2.9 cm) draw the circumcircle through A, B and C. (c) The locus of points equidistant from AB and BC is the bisector of ∠ ABC. Draw it; it passes through O and meets the circumcircle again at D (the point diametrically opposite B). (d) BD is a diameter and the perpendicular bisector of AC, so AB = BC and AD = DC (with ∠ BAD = ∠ BCD = 90^°, ∠ ADC = 120^°). ABCD is a kite (a cyclic kite).

ICSE 2025 board paper · 1 mark

Statement 1: The point which is equidistant from three non-collinear points D, E and F is the circumcentre of the ΔDEF. Statement 2: The incentre of a triangle is the point where the bisector of the angles intersects.

Both the statements are true.
Both the statements are false.
Statement 1 is true, and Statement 2 is false.
Statement 1 is false, and Statement 2 is true.

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Both the statements are true.

ICSE 2025 compartment paper · 1 mark

The locus of a toy bird fixed at the tip of one of the blades of a rotating ceiling fan is a:

straight line
circle
semi-circular arc
diameter of the circle so formed

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circle

ICSE 2025 board paper · 4 marks

Use ruler and compass for the following construction. Construct a ΔABC, where AB = 6 cm, AC = 4.5 cm and ∠BAC = 120°. Construct a circle circumscribing the ΔABC. Measure and write down the length of the radius of the circle.

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Steps: (1) Draw AB = 6 cm. (2) At A construct ∠ BAC = 120^° and cut AC = 4.5 cm; join BC. (3) Draw the perpendicular bisectors of AB and AC; they meet at O (outside the triangle, since the triangle is obtuse). (4) With centre O and radius OA, draw the circle; it passes through A, B and C. Radius of the circumcircle ≈ 5.3 cm. (Check: BC² = 6² + 4.5² - 2(6)(4.5)cos 120^° = 83.25, BC ≈ 9.12 cm; R = (BC)/(2 sin 120^°) ≈ 5.27 cm.)

ICSE 2025 compartment paper · 4 marks

Construct a regular hexagon of side 4.5 cm. Hence, construct a circle circumscribing the regular hexagon. Use ruler and compass for the construction. Measure and write down the radius of the circle.

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Steps of construction: 1. Draw AB = 4.5 cm. 2. At A and B construct angles of 120° (each interior angle of a regular hexagon is 120°) and cut off AF = BC = 4.5 cm. 3. Similarly, at C and F construct 120° angles and cut off CD = FE = 4.5 cm; join DE. ABCDEF is the regular hexagon. 4. Draw the perpendicular bisectors of AB and BC; they meet at O. 5. With centre O and radius OA, draw the circle; it passes through all six vertices. Radius of the circumscribing circle = 4.5 cm (in a regular hexagon each side subtends 60° at the centre, so the circumradius equals the side).

ICSE 2025 sample paper · 4 marks

Use ruler and compasses for the following question taking a scale of 10 m=1 cm. A park in a city is bounded by straight fences AB, BC, CD and DA. Given that AB = 50 m, BC = 63 m, ∠ABC = 75°. D is a point equidistant from the fences AB and BC. If ∠BAD = 90°, construct the outline of the park ABCD. Also locate a point P on the line BD for the flag post which is equidistant from the corners of the park A and B.

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Scale 10 m = 1 cm, so AB = 5 cm and BC = 6.3 cm. Steps of construction: 1. Draw AB = 5 cm. 2. At B construct ∠ABC = 75° and cut BC = 6.3 cm. 3. Construct the bisector of ∠ABC (the locus of points equidistant from the fences AB and BC). 4. At A construct ∠BAD = 90°; this perpendicular meets the bisector of ∠ABC at D. 5. Join CD. ABCD is the outline of the park. 6. Construct the perpendicular bisector of AB (the locus of points equidistant from A and B). It meets BD at P, the position of the flag post. (Check: ∠ABD = 37.5°, so AD = 5 tan 37.5° ≈ 3.8 cm, i.e. about 38 m; PA = PB = 2.5 ÷ cos 37.5° ≈ 3.2 cm, i.e. about 32 m.)