15 multiple-choice questions and 2 case-based questions on the current (rationalised) syllabus, in the style of the board paper's Sections A and E. Try each one first; the answer, a one-line reason and a worked solution open on a tap.
Choose one option. Section A of the board paper has 18 MCQs of 1 mark each, spread over all chapters.
A point \(P\) is \(13\) cm from the centre of a circle of radius \(5\) cm. The length of the tangent from \(P\) to the circle is
Why: The radius to the point of contact is perpendicular to the tangent: 13² − 5² = 144.
\(\triangle OTP\) is right-angled at the point of contact \(T\): \(PT = \sqrt{13^2 - 5^2} = \sqrt{144} = 12\) cm.
From a point \(P\) outside a circle, tangents are drawn to the circle. Which statement is correct?
Why: From an external point there are exactly two tangents, and their lengths are equal.
From a point outside a circle exactly two tangents can be drawn (and they are equal in length).
A line that meets a circle at exactly one point is called
Why: A secant meets the circle twice; a tangent touches it once.
A secant cuts the circle at two points; a tangent meets it at exactly one point, the point of contact.
\(PA\) and \(PB\) are tangents from \(P\) to a circle with centre \(O\). If \(\angle APB = 70^\circ\), then \(\angle AOB\) is
Why: In quadrilateral OAPB the angles at A and B are 90°, so ∠AOB = 180° − ∠APB.
\(\angle OAP = \angle OBP = 90^\circ\). Angle sum of \(OAPB\): \(\angle AOB = 360^\circ - 90^\circ - 90^\circ - 70^\circ = 110^\circ\).
A quadrilateral \(ABCD\) is drawn to circumscribe a circle. If \(AB = 6\) cm, \(BC = 7\) cm and \(CD = 5\) cm, then \(DA\) is
Why: Equal tangents give AB + CD = BC + DA.
Using equal tangent lengths from each vertex, \(AB + CD = BC + DA\): \(6 + 5 = 7 + DA \Rightarrow DA = 4\) cm.
\(PQ\) is a tangent at \(Q\) to a circle with centre \(O\) and radius \(9\) cm. If \(PQ = 12\) cm, then \(OP\) is
Why: OQ ⊥ PQ, so OP² = 9² + 12².
\(OP = \sqrt{9^2 + 12^2} = \sqrt{225} = 15\) cm.
Tangents \(PA\) and \(PB\) from \(P\) to a circle with centre \(O\) make \(\angle APB = 60^\circ\). If \(PA = 6\sqrt3\) cm, the radius of the circle is
Why: OP bisects ∠APB, so in right △OAP, tan 30° = OA/PA.
\(\angle APO = 30^\circ\) and \(\angle OAP = 90^\circ\). \(OA = PA\tan 30^\circ = 6\sqrt3 \times \dfrac{1}{\sqrt3} = 6\) cm.
The angle between a tangent to a circle and the radius drawn to the point of contact is
Why: Theorem: the tangent is perpendicular to the radius at the point of contact.
The tangent at any point of a circle is perpendicular to the radius through the point of contact, so the angle is \(90^\circ\).
A chord of the larger of two concentric circles, of radii \(25\) cm and \(7\) cm, touches the smaller circle. Its length is
Why: The chord is a tangent to the inner circle, so the radius 7 bisects it at right angles: half-chord = √(625 − 49).
The radius of the small circle meets the chord at right angles at its mid-point. Half-chord \(= \sqrt{25^2 - 7^2} = \sqrt{576} = 24\) cm, so the chord is \(48\) cm.
\(PA\) and \(PB\) are tangents from \(P\) to a circle. If \(\angle PAB = 65^\circ\), then \(\angle APB\) is
Why: PA = PB, so △PAB is isosceles with ∠PBA = 65° as well.
\(PA = PB \Rightarrow \angle PBA = \angle PAB = 65^\circ\). \(\angle APB = 180^\circ - 130^\circ = 50^\circ\).
\(PQ\) is a tangent at \(Q\) to a circle with centre \(O\) and radius \(5\) cm. If \(\angle POQ = 60^\circ\), then \(PQ\) is
Why: △OQP is right-angled at Q: PQ = OQ tan 60°.
\(\angle OQP = 90^\circ\), so \(PQ = OQ\tan 60^\circ = 5\sqrt3\) cm.
A circle is inscribed in \(\triangle ABC\) with \(AB = 8\) cm, \(BC = 10\) cm and \(CA = 12\) cm. The length of the tangent from \(A\) to the circle is
Why: With tangent lengths x, y, z from A, B, C: x + y = 8, y + z = 10, z + x = 12.
Adding: \(2(x + y + z) = 30\), so \(x + y + z = 15\) and \(x = 15 - (y + z) = 15 - 10 = 5\) cm.
\(PA\) and \(PB\) are tangents from \(P\) to a circle with centre \(O\). If \(\angle POA = 55^\circ\), then the angle between the two tangents, \(\angle APB\), is
Why: In right △OAP, ∠APO = 90° − 55° = 35°, and OP bisects ∠APB.
\(\angle OAP = 90^\circ\), so \(\angle APO = 90^\circ - 55^\circ = 35^\circ\). \(OP\) bisects \(\angle APB\) (congruent \(\triangle OAP\), \(\triangle OBP\)), so \(\angle APB = 70^\circ\).
\(PA\) and \(PB\) are tangents from an external point \(P\) to a circle, with \(PA = (3x - 2)\) cm and \(PB = (x + 6)\) cm. The length \(PA\) is
Why: Tangents from an external point are equal: 3x − 2 = x + 6.
\(3x - 2 = x + 6 \Rightarrow x = 4\), so \(PA = 3(4) - 2 = 10\) cm.
A circle is inscribed in a triangle whose sides are \(6\) cm, \(8\) cm and \(10\) cm (a right triangle). The radius of the circle is
Why: At the right angle the two tangent lengths equal r, so (6 − r) + (8 − r) = 10.
The radii to the two legs and the legs form a square of side \(r\) at the right angle. Tangent lengths: \(6 - r\) and \(8 - r\) along the legs, and these add to the hypotenuse: \(14 - 2r = 10 \Rightarrow r = 2\) cm.
Section E of the board paper has three case-based questions of 4 marks: a real-life passage, then parts of 1, 1 and 2 marks, with a choice (OR) on the 2-mark part.
A circular roundabout has centre \(O\) and radius \(15\) m. A straight road touches the roundabout at \(A\), and a traffic post \(P\) stands on this road \(20\) m from \(A\). A second straight road from \(P\) touches the roundabout at \(B\).
A circular plate fits exactly inside a triangular tray \(ABC\), touching the sides \(BC\), \(CA\) and \(AB\) at \(D\), \(E\) and \(F\) respectively. A student measures \(AF = 4\) cm, \(BD = 6\) cm and \(CE = 8\) cm.
This free set is separate from the chapter's question bank. On the Circles chapter page the revision notes and the first step of the Route to 95 are free for everyone. CBSE Essentials adds all 38 questions in the chapter bank (short and long answers, assertion–reason and more case studies) with their full step-marking schemes, the Route to 95 checkpoints with your progress saved, Skill Builders and the full common-mistakes library. There is no AI marking on CBSE Math Revision: you check your work against the marking scheme.
Original questions written by CBSE Math Revision for the CBSE 2026-27 syllabus; not taken from NCERT, NCERT Exemplar or CBSE papers. Every answer was re-checked by computer algebra and by an independent reviewer. CBSE Math Revision is independent and not affiliated with CBSE or NCERT. Spotted a slip? Tell us and it goes in the corrections log.