Circles Worksheet

(c) 90°

1. AB subtends 180° at the centre.
2. The angle at C on the remaining circle is half this central angle, namely 90°.

(b) Cyclic

1. 70° + 110° = 180°.
2. The converse of the cyclic-quadrilateral angle theorem gives cyclicity.

(b) Closer to the centre

1. For radius r and perpendicular distance d, half-chord length is √(r² − d²).
2. A larger chord requires a smaller d.

(b) 68°

1. Opposite interior angles of a cyclic quadrilateral sum to 180°.
2. 180° − 112° = 68°.

24 cm.

1. The perpendicular from the centre bisects the chord.
2. Half the chord, the distance 5 and radius 13 form a right triangle.
3. Half-chord = √(13² − 5²) = √144 = 12 cm; the full chord is 24 cm.

Distance 3 cm; the perpendicular bisects AB.

1. Let M be the perpendicular foot. Right triangles OMA and OMB have equal hypotenuses OA = OB and common leg OM.
2. RHS gives AM = MB, so each half of the chord is 4 cm.
3. OM² = OA² − AM² = 25 − 16 = 9; OM = 3 cm.
4. Conversely, if M is the midpoint, SSS makes the adjacent angles at M equal and supplementary, hence each is 90°.

Intersect two perpendicular bisectors; the circumcentre is inside, at the hypotenuse midpoint, or outside respectively.

1. Construct perpendicular bisectors of AB and BC. Non-collinearity ensures they meet at a unique O.
2. O is equidistant from A and B, and from B and C, so OA = OB = OC.
3. Draw the circle of radius OA. Any centre of a circle through all three points must lie on both bisectors, proving uniqueness.
4. For an acute triangle O is inside; for a right triangle it is the hypotenuse midpoint; for an obtuse triangle it is outside.

A diameter has maximum chord length 2r; circular wheels and an Ashoka Chakra motif are examples.

1. A circle consists of points at fixed positive distance r from its centre. A radius joins centre to circle; a chord joins two points on it; a diameter is a chord through the centre.
2. An arc is part of the circumference. A sector is bounded by two radii and an arc; a segment by a chord and an arc.
3. For chord AB, AB ≤ AO + OB = 2r by the triangle inequality. Equality occurs for a diameter with O between the endpoints.
4. A circular wheel maintains a constant axle height on level ground. Cultural examples include Ashoka Chakra, Dharmachakra and Sudarshan Chakra motifs.

Equal chords give equal central angles and distances, with converses true. The specified inscribed angle is 50°.

1. For equal chords AB and CD, triangles AOB and COD have equal radii and equal third sides; SSS gives equal central angles.
2. Conversely equal central angles and the two pairs of radii give equal chords by SAS.
3. Perpendiculars from O bisect the chords. Equal half-chords and equal radii give equal distances using right triangles.
4. Conversely equal perpendicular distances and equal radii give equal half-chords, hence equal full chords.
5. An inscribed angle with vertex on the remaining major arc is half the 100° central angle, so it is 50°. Angles in that same segment are equal.

∠A + ∠C = 180°; ∠C = 102°. A simple quadrilateral with supplementary opposite angles is cyclic.

1. The arcs intercepted by the opposite angles together make a complete 360° circle.
2. Each inscribed angle is half the measure of its intercepted arc.
3. Adding the two angles therefore gives 360°/2 = 180°.
4. The converse says a simple quadrilateral with supplementary opposite angles has its four vertices on one circle.
5. With ∠A = 78°, ∠C = 180° − 78° = 102°. The supplementary-angle criterion is an exact concyclicity test, not an inference from the drawing.

Circumcentre (3,4); radius 5 units.

1. The angle at (0,0) is right. The circle on the hypotenuse as diameter passes through the third vertex by the semicircle theorem and its converse.
2. Its centre is the midpoint of (6,0) and (0,8), namely (3,4).
3. Distances to the three vertices are √(3² + 4²) = 5, √((6 − 3)² + 4²) = 5 and √(3² + (8 − 4)²) = 5.
4. Equal distances verify both the centre and radius.