CBSE Class 10 Maths Circles Previous Year Questions

Join OQ and OS. Since OQ and OS are radii and PQ, PS are tangents at Q and S:
$\angle OQP = \angle OSP = 90°$ (tangent perpendicular to radius at point of contact)

In quadrilateral PQOS (sum of angles $= 360°$): $$\angle QPS + \angle OQP + \angle OSP + \angle QOS = 360°$$ $$50° + 90° + 90° + \angle QOS = 360°$$ $$\angle QOS = 130°$$ Since $\angle QRS$ is the angle subtended at a point R on the major arc by the same arc QS: $$\angle QRS = \frac{1}{2} \angle QOS = \frac{130°}{2} = \mathbf{65°}$$

Reflex $\angle AOB + $ minor $\angle AOB = 360°$ (complete angle)
Minor $\angle AOB = 360° – 260° = 100°$

Since OA $\perp$ LN and OB $\perp$ NM (radius $\perp$ tangent at point of tangency):
$\angle OAN = 90°$ and $\angle OBN = 90°$

In quadrilateral NAOB (sum of angles $= 360°$): $$\angle OAN + \angle AOB + \angle OBN + \angle ANB = 360°$$ $$90° + 100° + 90° + \angle ANB = 360°$$ $$\angle ANB = 360° – 280° = \mathbf{80°}$$

Infinite number of circles can be drawn.

Justification: $\angle PRQ = 90°$ (angle in a semicircle). The radius of any circle tangent to QR at R must be perpendicular to QR at R. Therefore the radius must lie along the line PR extended. Infinite circles with radii lying along the extension of PR and with R as the point of tangency can be drawn, one for each possible radius length.

Since $KM = ML$ (equal radii), $\angle MKL = \angle MLK = x$ (angles opposite equal sides).
By angle sum property in $\triangle KML$: $\theta + x + x = 180° \Rightarrow x = 90° – \dfrac{\theta}{2}$

Since $\angle MKP + \angle MKL = 180°$ (angles on a straight line): $$\angle MKP = 90° + \frac{\theta}{2} \quad \text{and similarly} \quad \angle MLQ = 90° + \frac{\theta}{2}$$
(i) When $\theta = 0°$: K and L coincide, and $\angle MKP = \angle MLQ = 90°$.
Therefore PQ is tangent to the circle, since the radius is perpendicular to PQ at the point of contact. ✓

(ii) The point of contact is K (or L), since $\angle MKP = 90°$.

In $\triangle QOR$: $OQ = OR$ (equal radii), so $\angle OQR = \angle ORQ = x$ (angles opposite equal sides).

Now $\angle PRO = 90°$ (radius perpendicular to tangent at point of contact).
$\angle PRO = \angle PRQ + \angle ORQ \Rightarrow 90° = \angle PRQ + x \Rightarrow \angle PRQ = (90° – x)$

Similarly, $\angle PQR = (90° – x)$

In $\triangle PQR$: $\angle PQR = \angle PRQ$, so the sides opposite these equal angles are equal: $$PR = PQ$$ Hence, the tangents PQ and PR are equal in length. $\blacksquare$

$\angle ACP + \angle ACB = 180°$ (straight line), so $\angle ACP = 110°$.

In $\triangle APC$ (angle sum property): $$\angle CAP = 180° – 110° – 30° = 40°$$ $\angle OAP = 90°$ (tangent perpendicular to radius). So $\angle OAC = 90° – 40° = 50°$.

Join OC. In $\triangle OAC$: $OA = OC$ (equal radii), so $\angle OCA = \angle OAC = 50°$.
$\angle OCB = 70° – 50° = 20°$.

In $\triangle OCB$: $OB = OC$ (equal radii), so $\angle OBC = \angle OCB = \mathbf{20°}$.

Since PQ and PT are tangents from external point P: $PQ = PT = 28$ cm.

Let $LQ = x$, so $PL = (28 – x)$ cm. Let $MT = y$, so $PM = (28 – y)$ cm.
By equal tangents: $LN = LQ = x$ and $NM = MT = y$, so $LM = x + y$.

Perimeter of $\triangle PLM = PL + LM + PM$ $$= (28 – x) + (x + y) + (28 – y) = 28 + 28 = \mathbf{56} \text{ cm}$$

$\angle OPQ = 90°$ (tangent perpendicular to radius).
Let $\angle PQO = x$. Since $\triangle OPQ$ is isosceles, $\angle QOP = x$ as well.

By angle sum property in $\triangle OPQ$: $$90° + x + x = 180° \Rightarrow 2x = 90° \Rightarrow x = 45°$$ Therefore, $\angle OQP = \mathbf{45°}$.

AP and AR are tangents from A, so $AP = AR = 4$ cm.
PB and BQ are tangents from B, so $BP = BQ = 3$ cm.
$CR = AC – AR = 11 – 4 = 7$ cm. CR and CQ are tangents from C, so $CR = CQ = 7$ cm.

$$BC = BQ + CQ = 3 + 7 = \mathbf{10} \text{ cm}$$

The statement is TRUE.

Justification: Since tangents are drawn at A and E (the endpoints of the diameter), $\angle OAB = \angle OED = 90°$ (radius perpendicular to tangent at point of contact). These are co-interior angles (same-side interior angles) formed by the transversal AE with lines AB and ED. Since they are supplementary ($90° + 90° = 180°$), it follows that $AB \| ED$. Hence, at least one pair of opposite sides of AEDB is parallel.

Given: PA and PB are tangents from external point P to a circle with centre O.
To Prove: OP bisects $\angle APB$, i.e., $\angle OPA = \angle OPB$.

Proof:
In $\triangle OAP$ and $\triangle OBP$:
 • $OA = OB$ (radii of the same circle)
 • $PA = PB$ (tangents from an external point are equal)
 • $\angle OAP = \angle OBP = 90°$ (radius perpendicular to tangent)

By RHS congruence: $\triangle OAP \cong \triangle OBP$

Therefore by CPCT: $\angle OPA = \angle OPB$

Hence, OP bisects $\angle APB$. $\blacksquare$

Since $OP = OR$ (radii): $\angle OPR = \angle ORP$  …(i)

By exterior angle property: $\angle QOR = \angle OPR + \angle ORP$  …(ii)
Also: $\angle QOR = \angle QOS + \angle ROS$  …(iii)

From (ii) and (iii): $\angle OPR + \angle ORP = \angle QOS + \angle ROS$
Using (i): $2\angle ORP = \angle QOS + \angle ROS$, so $\angle ROS = \angle QOS$.

In $\triangle ORS$ and $\triangle OQS$:
 • $OS = OS$ (common)
 • $OR = OQ$ (radii)
 • $\angle ROS = \angle QOS$ (proved above)

By SAS congruence: $\triangle ORS \cong \triangle OQS$
So $\angle ORS = \angle OQS$ by CPCT.

Since RS is a tangent to the circle, $\angle ORS = 90°$. Therefore $\angle OQS = 90°$, which means OS is perpendicular to the tangent at Q, so OS is also a tangent to the circle. $\blacksquare$

Given: ME and NE are tangents from external point E to a circle with centre O.
To Prove: $\angle MEO = \angle NEO$
Construction: Join OE.

In $\triangle OME$ and $\triangle ONE$:
 • $OM = ON$ (radii)
 • $OE = OE$ (common)
 • $ME = NE$ (tangents from an external point are equal in length)

By SSS congruence: $\triangle OME \cong \triangle ONE$

Therefore by CPCT: $\angle MEO = \angle NEO$

Hence OE bisects $\angle MEN$. $\blacksquare$

Let $\angle TPQ = \theta$.
$\angle TPO = 90°$ (tangent perpendicular to radius), so $\angle OPQ = 90° – \theta$.

Since $TP = TQ$ (tangents from external point are equal): $\angle TPQ = \angle TQP = \theta$.

In $\triangle PQT$ (angle sum property): $$\angle TPQ + \angle TQP + \angle PTQ = 180°$$ $$\theta + \theta + \angle PTQ = 180°$$ $$\angle PTQ = 180° – 2\theta = 2(90° – \theta) = 2\angle OPQ$$ Hence Proved. $\blacksquare$

$\angle OAB = 30°$ and $\angle OAP = 90°$ (tangent perpendicular to radius).
So $\angle PAB = 90° – 30° = 60°$.

Since $AP = BP$ (tangents from P are equal): $\angle PAB = \angle PBA = 60°$.
In $\triangle ABP$: $\angle APB = 180° – 60° – 60° = 60°$.
Therefore $\triangle ABP$ is equilateral, so $PA = PB = AB$: $$\boxed{PA = 6 \text{ cm}}$$ In right $\triangle OAP$, $\angle OPA = 30°$: $$\tan 30° = \frac{OA}{AP} \Rightarrow \frac{1}{\sqrt{3}} = \frac{OA}{6} \Rightarrow OA = \frac{6}{\sqrt{3}} = \boxed{2\sqrt{3} \text{ cm}}$$

Given: Let ABCD be a parallelogram circumscribing a circle, touching sides AB, BC, CD, DA at points P, Q, R, S respectively.

Since tangents from an external point are equal:
$AP = AS$, $BP = BQ$, $CR = CQ$, $DR = DS$

Adding all: $$(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)$$ $$AB + CD = AD + BC$$ Since ABCD is a parallelogram: $AB = CD$ and $AD = BC$.
Substituting: $2AB = 2AD \Rightarrow AB = AD$

Since $AB = AD$ and opposite sides of a parallelogram are equal, all four sides are equal: $$AB = BC = CD = DA$$ A parallelogram with all sides equal is a rhombus. $\blacksquare$

TP and TQ are tangents from T, so $\triangle TPQ$ is isosceles. OT bisects PQ at R.
$PR = RQ = 4$ cm.

$OR = \sqrt{OP^2 – PR^2} = \sqrt{25 – 16} = 3$ cm.

$\angle TPR + \angle RPO = 90°$ (since $\angle TPO = 90°$)  …(i)
$\angle TPR + \angle PTR = 90°$ (angle sum in right $\triangle TPR$)  …(ii)
From (i) and (ii): $\angle RPO = \angle PTR$.

Therefore right $\triangle TRP \sim$ right $\triangle PRO$ (by AA): $$\frac{TP}{PO} = \frac{RP}{RO} \Rightarrow \frac{TP}{5} = \frac{4}{3} \Rightarrow TP = \frac{20}{3} \approx \mathbf{6.67} \text{ cm}$$

Given: PA and PB are tangents from external point P to a circle with centre O (radius $r$).
To Prove: $PA = PB$

Draw radii OA and OB. Since the radius is perpendicular to the tangent at the point of contact: $$\angle OAP = \angle OBP = 90°$$ In $\triangle OAP$ and $\triangle OBP$:
 • $OA = OB$ (radii of the same circle)
 • $OP = OP$ (common hypotenuse)
 • $\angle OAP = \angle OBP = 90°$

By RHS congruence: $\triangle OAP \cong \triangle OBP$

Therefore by CPCT: $\mathbf{PA = PB}$ $\blacksquare$

Given: Quadrilateral ABCD circumscribes a circle with centre O. Sides AB, BC, CD, DA touch the circle at P, Q, R, S respectively.
To Prove: $\angle AOB + \angle COD = 180°$ and $\angle DOA + \angle BOC = 180°$

Join OP, OQ, OR, OS. Since tangents from an external point make equal angles with the line to the centre:
$\angle AOP = \angle AOS$, $\angle BOP = \angle BOQ$, $\angle COQ = \angle COR$, $\angle DOR = \angle DOS$

The sum of all eight angles equals $360°$: $$2(\angle AOP + \angle BOP + \angle COQ + \angle DOR) = 360°$$ $$(\angle AOB) + (\angle COD) = 180°$$ Similarly: $$(\angle BOC) + (\angle DOA) = 360° – 180° = 180°$$ Hence, opposite sides of the circumscribed quadrilateral subtend supplementary angles at the centre. $\blacksquare$

1. Answer: (C) $150°$
$\angle ORP = \angle OQP = 90°$ (radius perpendicular to tangent). In quadrilateral with $\angle QPR = 30°$: $$\angle ROQ = 360° – 90° – 90° – 30° = 150°$$
2. Answer: (A) $75°$
In $\triangle OQR$: $\angle OQR = \angle ORQ = \dfrac{180° – 150°}{2} = 15°$
$\angle RQP = \angle OQP – \angle OQR = 90° – 15° = 75°$

3. Answer: (B) $75°$
$\angle RSQ = \dfrac{1}{2}\angle ROQ = \dfrac{150°}{2} = 75°$ (angle at centre is double the angle at circumference)

4. Answer: (A) $90°$
$\angle ORP = 90°$ because the radius is perpendicular to the tangent at the point of contact.

5. Answer: (B) $180°$
$\angle ORP + \angle OQP = 90° + 90° = 180°$

(i) Finding AD and BE:
Let $AD = x$. By equal tangents: $AF = x$, $BD = BE = (12 – x)$, $CF = CE = (10 – x)$.
Since $BE + CE = BC = 8$: $(12 – x) + (10 – x) = 8 \Rightarrow 22 – 2x = 8 \Rightarrow x = 7$
$$AD = 7 \text{ cm}, \quad BE = 12 – 7 = \mathbf{5} \text{ cm}$$ OR Area of $\triangle OAB = \dfrac{1}{2} \times OD \times AB = \dfrac{1}{2} \times 4 \times 12 = \mathbf{24 \text{ cm}^2}$

(ii) Perimeter of $\triangle ABC$: $$AB + BC + CA = 12 + 8 + 10 = \mathbf{30} \text{ cm}$$ (iii) Length of CF: $$CF = AC – AF = 10 – 7 = \mathbf{3} \text{ cm}$$