Circles — Important Questions

90°

Step 1: We have a semicircle with diameter AB and point C on the semicircle. Step 2: The arc AB (semicircle) subtends 180° at center O. Step 3: By Theorem 9.7, angle at circumference = (1/2) × central angle. Step 4: Therefore, ∠ACB = 180°/2 = 90°. This is known as 'angle in a semicircle is a right angle' - a fundamental property of circles.

They are concyclic

Step 1: We have chord PQ subtending equal angles at points R and S on same side, so ∠PRQ = ∠PSQ. Step 2: By Theorem 9.9, if a line segment subtends equal angles at two points on same side, all four points lie on a circle. Step 3: This means points P, Q, R, and S are concyclic (lie on the same circle). Step 4: The converse is also true - if points are concyclic, they subtend equal angles (angles in same segment are equal).

110°

Step 1: ABCD is a cyclic quadrilateral with ∠A = 70°, and we need to find ∠C. Step 2: By Theorem 9.10, opposite angles of a cyclic quadrilateral are supplementary (sum = 180°). Step 3: Since A and C are opposite angles: ∠A + ∠C = 180°. Step 4: Therefore, ∠C = 180° - 70° = 110°. This property helps identify cyclic quadrilaterals.

5 cm

Step 1: Let chord AB = 6 cm, distance from center O to chord = 4 cm, perpendicular meets chord at M. Step 2: By perpendicular bisector property, AM = MB = 6/2 = 3 cm. Step 3: In right triangle OAM: OM = 4 cm, AM = 3 cm, OA = radius. Step 4: Using Pythagoras theorem: OA² = OM² + AM² = 4² + 3² = 16 + 9 = 25. Therefore, OA = 5 cm.