Circles

Circles

Circles carries 7 marks in the board paper. It got exactly 7 in the 2026-27 sample paper and in the 2025 and both 2026 board papers.

The marks usually come as one or two 1-mark MCQs (angles between tangents, a tangent length by Pythagoras), a 2-mark question, and a 3-mark proof: one of the two theorems, a quadrilateral or parallelogram circumscribing a circle, or parallel tangents with $\angle POQ = 90^\circ$. The sample paper had a case study on tangents to a garden, and the Feb 2026 paper set a 5-mark question split into 1 + 1 + 1 + 2.

Where marks are usually lost:

• not writing the reason "radius $\perp$ tangent" or "tangents from an external point are equal" beside the step that uses it;
• theorem proofs without a figure, "Given", "To prove" and "Construction";
• mixing up $\angle APB$ (at the external point) with $\angle AOB$ (at the centre);
• using Pythagoras in a triangle whose right angle has not been justified.

Revise in 5 Minutes

Basics

• A tangent meets the circle at exactly one point; a secant at two.
• Tangents from a point inside / on / outside the circle: 0 / 1 / 2.

The two theorems (proofs are asked)

Results that follow (tangents $PA$, $PB$ from $P$, centre $O$, radius $r$)

• $PA^2 = OP^2 - r^2$
• $OP$ bisects $\angle APB$ and $\angle AOB$, and $OP$ is the perpendicular bisector of $AB$.
• $\angle APB + \angle AOB = 180^\circ$
• $\angle PTQ = 2\angle OPQ$ for tangents $TP$, $TQ$.
• Tangents at the ends of a diameter are parallel, a diameter apart.
• Concentric circles: the chord of the larger circle touching the smaller one is bisected there; half-chord $= \sqrt{R^2 - r^2}$.
• Quadrilateral $ABCD$ circumscribing a circle: $AB + CD = AD + BC$; opposite sides subtend supplementary angles at the centre.
• A parallelogram circumscribing a circle is a rhombus.
• Parallel tangents $XY$, $X'Y'$ cut by a third tangent at $A$ and $B$: $\angle AOB = 90^\circ$.
• A third tangent at $C$ meeting $PA$, $PB$ at $X$, $Y$: perimeter of $\triangle PXY = 2PA$.
• Circle inscribed in $\triangle ABC$ (touching $BC$, $CA$, $AB$ at $D$, $E$, $F$): $AF = AE = s - BC$, where $s$ is half the perimeter; in a right triangle with legs $a$, $b$ and hypotenuse $c$, $r = \frac{a + b - c}{2}$.
• Angle chases: tangent $\perp$ diameter at its end, and the angle in a semicircle is $90^\circ$.

Traps

• Write the reason with every use of $90^\circ$ at a point of contact.
• $OP$, not the tangent, is the hypotenuse.
• In a theorem proof, write Given, To prove, Construction and Proof, with a figure.

How to use this page: try each question on paper first, then read the answer. The marks against each step show what an examiner looks for. The 1-mark MCQs and Assertion-Reason questions are in the quiz at the end, together with questions that test how well you understand the chapter; every quiz answer comes with its explanation.

Short Answer Questions (2 and 3 Marks)

Question 1 (2 marks)

Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Answer.

Model answer:

Let $AB$ be a diameter of a circle with centre $O$. Let $PQ$ be the tangent at $A$ and $RS$ the tangent at $B$, with $P$ and $S$ on opposite sides of $AB$.

The tangent at any point is perpendicular to the radius through the point of contact. So $OA \perp PQ$ and $OB \perp RS$, which gives $\angle PAB = 90^\circ$ and $\angle ABS = 90^\circ$.

$\angle PAB$ and $\angle ABS$ are alternate angles made by the transversal $AB$ with the lines $PQ$ and $RS$, and they are equal. Hence $PQ \parallel RS$.

Marking scheme:

1. Let $AB$ be a diameter with centre $O$, and $PQ$, $RS$ the tangents at $A$ and $B$; the radius is perpendicular to the tangent, so $\angle OAP = 90^\circ$ and $\angle OBS = 90^\circ$ — 1 mark
2. So $\angle PAB = \angle ABS = 90^\circ$; these are alternate angles made by the transversal $AB$, hence $PQ \parallel RS$ — 1 mark

Question 2 (2 marks)

A circular pond in a park in Bhopal has a radius of 9 m. Anu stands at a point $P$, 15 m from the centre of the pond. She walks along a straight path that just touches the edge of the pond at $T$. How far does she walk to reach $T$?

Answer.

1. $PT$ is a tangent, so $OT \perp PT$ and $\triangle OTP$ is right-angled at $T$ — 1 mark
2. $PT = \sqrt{15^2 - 9^2} = \sqrt{144} = 12$ m — 1 mark

Question 3 (2 marks)

In the figure, $TP$ is a tangent from $T$ to a circle with centre $M$ and radius 12 cm, touching it at $P$, and $TP = 16$ cm. The same line $TP$ also touches a smaller circle with centre $N$ and radius 3 cm at $A$, where $N$ lies on $TM$. Find $TA$ and $AP$.

Answer.

1. $MP \perp TP$ and $NA \perp TP$ (radius and tangent), and $\angle T$ is common, so $\triangle TAN \sim \triangle TPM$ (AA) — 1 mark
2. $\frac{TA}{TP} = \frac{NA}{MP}$, so $TA = 16 \times \frac{3}{12} = 4$ cm and $AP = 16 - 4 = 12$ cm — 1 mark

Question 4 (2 marks)

A quadrilateral $ABCD$ is drawn to circumscribe a circle. Its sides $AB$, $BC$, $CD$ and $DA$ touch the circle at $P$, $Q$, $R$ and $S$ respectively. Prove that $AB + CD = AD + BC$.

Answer.

1. Tangents from an external point are equal: $AP = AS$, $BP = BQ$, $CR = CQ$, $DR = DS$ — 1 mark
2. Adding: $(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)$, that is, $AB + CD = AD + BC$ — 1 mark

Question 5 (2 marks)

Two tangents $PA$ and $PB$ are drawn from an external point $P$ to a circle with centre $O$ such that $\angle APB = 120^\circ$. Prove that $OP = 2AP$.

Answer.

1. $OP$ bisects $\angle APB$, so $\angle APO = 60^\circ$; and $\angle OAP = 90^\circ$ (radius and tangent) — 1 mark
2. In right $\triangle OAP$: $\cos 60^\circ = \frac{AP}{OP}$, so $\frac{1}{2} = \frac{AP}{OP}$ and $OP = 2AP$ — 1 mark

Question 6 (3 marks)

Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

Answer.

Model answer:

Given: a circle with centre $O$ and a tangent $XY$ to the circle at the point $P$.

To prove: $OP \perp XY$.

Construction: take any point $Q$ on $XY$ other than $P$, and join $OQ$.

Proof: a tangent meets the circle at only one point, $P$. So $Q$ lies outside the circle. Let $OQ$ meet the circle at $R$. Then $OQ = OR + RQ$, so $OQ > OR$.

But $OR = OP$ (radii of the same circle), so $OQ > OP$.

This is true for every point $Q$ on $XY$ other than $P$. So $OP$ is the shortest of all the segments joining $O$ to points of $XY$.

The shortest segment from a point to a line is the perpendicular to the line. Hence $OP \perp XY$.

Marking scheme:

1. Given: circle with centre $O$, tangent $XY$ at $P$. To prove: $OP \perp XY$. Take any point $Q$ on $XY$ other than $P$ and join $OQ$ — 1 mark
2. $Q$ lies outside the circle (a tangent meets the circle only at $P$), so if $OQ$ meets the circle at $R$, then $OQ > OR = OP$ — 1 mark
3. So $OP$ is the shortest of all segments from $O$ to points of $XY$; the shortest segment from a point to a line is the perpendicular, hence $OP \perp XY$ — 1 mark

Question 7 (3 marks)

In the figure, $TP$ and $TQ$ are the two tangents to a circle with centre $O$ from an external point $T$. Prove that $\angle PTQ = 2\angle OPQ$.

Answer.

Model answer:

Let $\angle PTQ = \theta$.

$TP = TQ$, since the lengths of tangents from an external point are equal. So $\triangle TPQ$ is isosceles, and $\angle TPQ = \angle TQP = \frac{180^\circ - \theta}{2} = 90^\circ - \frac{\theta}{2}$.

The radius is perpendicular to the tangent at the point of contact, so $\angle OPT = 90^\circ$.

$\angle OPQ = \angle OPT - \angle TPQ = 90^\circ - \left(90^\circ - \frac{\theta}{2}\right) = \frac{\theta}{2} = \frac{1}{2}\angle PTQ$.

Hence $\angle PTQ = 2\angle OPQ$.

Marking scheme:

1. Let $\angle PTQ = \theta$; $TP = TQ$ (tangents from $T$), so $\angle TPQ = \angle TQP = \frac{180^\circ - \theta}{2} = 90^\circ - \frac{\theta}{2}$ — 1 mark
2. $\angle OPT = 90^\circ$ (radius and tangent), so $\angle OPQ = 90^\circ - \angle TPQ$ — 1 mark
3. $\angle OPQ = 90^\circ - (90^\circ - \frac{\theta}{2}) = \frac{\theta}{2}$, so $\angle PTQ = 2\angle OPQ$ — 1 mark

Question 8 (3 marks)

In the figure, $XY$ and $X'Y'$ are two parallel tangents to a circle with centre $O$. Another tangent $AB$, with point of contact $C$, meets $XY$ at $A$ and $X'Y'$ at $B$. Prove that $\angle AOB = 90^\circ$.

Answer.

Model answer:

Let $XY$ touch the circle at $P$ and $X'Y'$ touch it at $Q$. Join $OP$, $OC$, $OQ$, $OA$ and $OB$.

In right triangles $OPA$ and $OCA$: $OP = OC$ (radii), $OA$ is common and $\angle OPA = \angle OCA = 90^\circ$ (radius $\perp$ tangent). So $\triangle OPA \cong \triangle OCA$ (RHS), and $\angle OAP = \angle OAC$. Hence $\angle OAB = \frac{1}{2}\angle PAB$.

In the same way, $\triangle OQB \cong \triangle OCB$, so $\angle OBA = \frac{1}{2}\angle QBA$.

$XY \parallel X'Y'$ and $AB$ is a transversal, so $\angle PAB + \angle QBA = 180^\circ$ (co-interior angles).

Therefore $\angle OAB + \angle OBA = \frac{1}{2} \times 180^\circ = 90^\circ$, and in $\triangle AOB$, $\angle AOB = 180^\circ - 90^\circ = 90^\circ$.

Marking scheme:

1. Let $XY$ touch the circle at $P$ and $X'Y'$ at $Q$. $\triangle OPA \cong \triangle OCA$ (RHS: $OP = OC$, $OA$ common), so $\angle OAP = \angle OAC$, that is, $\angle OAB = \frac{1}{2}\angle PAB$ — 1 mark
2. Similarly $\angle OBA = \frac{1}{2}\angle QBA$; as $XY \parallel X'Y'$, $\angle PAB + \angle QBA = 180^\circ$ (co-interior angles) — 1 mark
3. So $\angle OAB + \angle OBA = 90^\circ$, and in $\triangle AOB$, $\angle AOB = 180^\circ - 90^\circ = 90^\circ$ — 1 mark

Question 9 (3 marks)

Prove that in two concentric circles, the chord of the larger circle which touches the smaller circle is bisected at the point of contact. Hence find the length of such a chord if the radii of the circles are 10 cm and 6 cm.

Answer.

Model answer:

Let two circles have the same centre $O$, and let the chord $AB$ of the larger circle touch the smaller circle at $P$. Join $OP$.

$AB$ is a tangent to the smaller circle at $P$, so $OP \perp AB$.

For the larger circle, $AB$ is a chord and $OP$ is the perpendicular from the centre to the chord. The perpendicular from the centre to a chord bisects it, so $AP = PB$. (Or: $\triangle OPA \cong \triangle OPB$ by RHS, since $OA = OB$, $OP$ is common and $\angle OPA = \angle OPB = 90^\circ$.)

With radii 10 cm and 6 cm: in right $\triangle OPA$, $AP = \sqrt{OA^2 - OP^2} = \sqrt{100 - 36} = 8$ cm. So $AB = 2 \times 8 = 16$ cm.

Marking scheme:

1. Let $AB$ touch the smaller circle at $P$; $OP \perp AB$ (radius and tangent) — 1 mark
2. $AB$ is a chord of the larger circle and $OP$ is the perpendicular from the centre to it, so $AP = PB$ — 1 mark
3. $AP = \sqrt{10^2 - 6^2} = 8$ cm, so $AB = 16$ cm — 1 mark

Long Answer and Case-Based Questions

Question 10 (5 marks)

In the figure, $PA$ and $PB$ are tangents from an external point $P$ to a circle with centre $O$, and the chord $AB$ meets $OP$ at $M$.

(i) Prove that the lengths of tangents drawn from an external point to a circle are equal, that is, $PA = PB$. (3 marks)

Answer.

Model answer:

Given: $PA$ and $PB$ are tangents from $P$ to a circle with centre $O$, touching it at $A$ and $B$.

To prove: $PA = PB$.

Construction: join $OA$, $OB$ and $OP$.

Proof: the tangent at any point is perpendicular to the radius through the point of contact, so $\angle OAP = \angle OBP = 90^\circ$.

In right triangles $OAP$ and $OBP$: $OA = OB$ (radii of the same circle) and the hypotenuse $OP$ is common.

So $\triangle OAP \cong \triangle OBP$ by RHS congruence. Hence $PA = PB$ (corresponding parts of congruent triangles).

Marking scheme:

1. Join $OA$, $OB$ and $OP$; $\angle OAP = \angle OBP = 90^\circ$ (radius is perpendicular to the tangent) — 1 mark
2. In $\triangle OAP$ and $\triangle OBP$: $OA = OB$ (radii), $OP$ is common — 1 mark
3. $\triangle OAP \cong \triangle OBP$ (RHS), so $PA = PB$ (CPCT) — 1 mark

(ii) Hence prove that $OP$ is the perpendicular bisector of the chord $AB$. (2 marks)

Answer.

1. From (i), $\angle APO = \angle BPO$; in $\triangle PAM$ and $\triangle PBM$: $PA = PB$, $\angle APM = \angle BPM$, $PM$ common, so $\triangle PAM \cong \triangle PBM$ (SAS) — 1 mark
2. So $AM = BM$ and $\angle AMP = \angle BMP$; as they add up to $180^\circ$, each is $90^\circ$; hence $OP$ is the perpendicular bisector of $AB$ — 1 mark

Question 11 (5 marks)

In the figure, a circle with centre $O$ is inscribed in $\triangle ABC$. It touches $BC$, $CA$ and $AB$ at $D$, $E$ and $F$ respectively.

(i) Prove that $OB$ bisects $\angle ABC$. (2 marks)

Answer.

1. $\angle OFB = \angle ODB = 90^\circ$ (radius $\perp$ tangent); in $\triangle OFB$ and $\triangle ODB$: $OF = OD$ (radii), $OB$ common — 1 mark
2. $\triangle OFB \cong \triangle ODB$ (RHS), so $\angle OBF = \angle OBD$; hence $OB$ bisects $\angle ABC$ — 1 mark

(ii) Hence prove that $\angle BOC = 90^\circ + \frac{1}{2}\angle A$. (2 marks)

Answer.

1. In the same way $OC$ bisects $\angle ACB$; so in $\triangle BOC$, $\angle BOC = 180^\circ - \frac{1}{2}\angle B - \frac{1}{2}\angle C$ — 1 mark
2. $\angle B + \angle C = 180^\circ - \angle A$, so $\angle BOC = 180^\circ - \frac{1}{2}(180^\circ - \angle A) = 90^\circ + \frac{1}{2}\angle A$ — 1 mark

(iii) If $\angle A = 70^\circ$, find $\angle BOC$. (1 mark)

Answer.

1. $\angle BOC = 90^\circ + 35^\circ = 125^\circ$ — 1 mark

Question 12 (4 marks)

A school in Mysuru has a garden in the shape of a right triangle $ABC$, with $\angle B = 90^\circ$, $AB = 20$ m, $BC = 21$ m and $AC = 29$ m. A circular flower bed with centre $O$ and radius $r$ is laid out so that it touches all three sides: $AB$ at $F$, $BC$ at $D$ and $CA$ at $E$, as shown in the figure.

(i) Name the tangent segments equal to $AF$ and to $CD$, and give the reason. (1 mark)

Answer.

1. $AF = AE$ and $CD = CE$, because tangents drawn from an external point to a circle are equal — 1 mark

(ii) Show that $BF = BD = r$, and write $AF$ and $CD$ in terms of $r$. (1 mark)

Answer.

1. $OFBD$ has three right angles ($\angle B$ and two radius-tangent angles) and $OF = OD = r$, so it is a square: $BF = BD = r$; $AF = 20 - r$, $CD = 21 - r$ — 1 mark

(iii) Using $AC = 29$ m, find the radius $r$ of the flower bed. (2 marks)

Answer.

1. $AC = AE + EC = AF + CD = (20 - r) + (21 - r)$ — 1 mark
2. $41 - 2r = 29$, so $r = 6$ m — 1 mark

OR

(iii) Taking $r = 6$ m, find the area of the garden and check that it equals $\frac{1}{2} \times r \times$ (perimeter of $\triangle ABC$). (2 marks)

Answer.

1. Area $= \frac{1}{2} \times 20 \times 21 = 210$ m$^2$ — 1 mark
2. $\frac{1}{2} \times 6 \times (20 + 21 + 29) = 3 \times 70 = 210$ m$^2$; the two are equal (area of $\triangle OAB$ + $\triangle OBC$ + $\triangle OCA$) — 1 mark

Question 13 (4 marks)

Neel's family pushes a round table of radius 60 cm into a corner $P$ of their room, where two walls meet at a right angle. The table top touches the two walls at $A$ and $B$. The figure shows the top view, with $O$ the centre of the table top.

(i) What kind of quadrilateral is $OAPB$? Give a reason. (1 mark)

Answer.

1. A square: $\angle OAP = \angle OBP = 90^\circ$ (radius and tangent), $\angle APB = 90^\circ$, so all angles are right angles, and $OA = OB$ (radii) — 1 mark

(ii) Find the distance $PA$ of the corner from the point where the table touches the wall. (1 mark)

Answer.

1. $OAPB$ is a square, so $PA = OB = 60$ cm — 1 mark

(iii) How far is the corner $P$ from the nearest point of the edge of the table? (Use $\sqrt{2} = 1.41$) (2 marks)

Answer.

1. $OP$ is the diagonal of the square: $OP = 60\sqrt{2} = 84.6$ cm — 1 mark
2. The nearest point of the edge lies on $OP$, so the distance $= 84.6 - 60 = 24.6$ cm — 1 mark

OR

(iii) Find the length of the segment $AB$ and the distance of $O$ from $AB$. (Use $\sqrt{2} = 1.41$) (2 marks)

Answer.

1. $AB$ is a diagonal of the square $OAPB$: $AB = 60\sqrt{2} = 84.6$ cm — 1 mark
2. The diagonals of a square bisect each other at right angles, so the distance of $O$ from $AB$ is half of $OP$: $30\sqrt{2} = 42.3$ cm — 1 mark