JEE Advanced Maths: Circles (Geometry) questions with solutions

Q1. A triangle ABC is inscribed in a circle with centre O, and the angle at A is 60 degrees. From O a perpendicular OD is dropped onto the chord BC. Find the measure of angle BOD.

1. 30 deg
2. 45 deg
3. 60 deg
4. 90 deg

Answer: 60 deg

The central angle BOC = 2 * angle A = 2 * 60 = 120 degrees (angle at centre is twice the angle at the circumference on the same arc). Since OD is perpendicular to chord BC, it bisects angle BOC, so angle BOD = 120/2 = 60 degrees.

Q2. In a circle, the chord ED is parallel to the diameter AC, and angle CBE = 50 degrees (B is a point on the circle). Find angle CED.

1. 40 degrees
2. 50 degrees
3. 30 degrees
4. 60 degrees

Answer: 40 degrees

Angle CBE and angle CAE subtend the same arc CE, so angle CAE = angle CBE = 50 degrees. Since AC is a diameter, the angle in the semicircle angle AEC = 90 degrees. In triangle AEC, angle ACE = 180 - 90 - 50 = 40 degrees. Because ED is parallel to AC, the alternate angle gives angle CED = angle ACE =... actually angle DEC = angle ECA (alternate angles with transversal EC) = 40 degrees. Hence angle CED = 40 degrees.

Q3. From an external point B, two lines BA and BC are drawn tangent to a circle, touching it at A and C respectively. E is a point on the major arc AC such that the inscribed angle AEC = 110 deg. Find the angle ABC.

1. 40 deg
2. 70 deg
3. 110 deg
4. 20 deg

Answer: 40 deg

Inscribed angle AEC = 110 deg subtends the minor arc AC, so the central angle AOC (reflex side) relates as: the arc not containing E has central angle = 2*110 = 220 deg, hence the central angle AOC on E's side = 360 - 220 = 140 deg. In quadrilateral OABC, OA perpendicular to BA and OC perpendicular to BC give angles 90 deg at A and C. Sum of quadrilateral angles = 360: angle ABC = 360 - 90 - 90 - 140 = 40 deg.

Q4. Two chords TY and OP of a circle intersect at point K such that TK = 2, KY = 16, and KP = 2*(KO). Find the length OP.

1. 8
2. 10
3. 12
4. 14

Answer: 12

By the intersecting chords theorem, TK*KY = OK*KP. Let KO = a, KP = 2a. Then 2*16 = a*2a => 32 = 2a² => a² = 16 => a = 4. So KO = 4, KP = 8, and OP = KO + KP = 4 + 8 = 12.