NCERT Solutions Class 10 Maths Chapter 10: Circles

#### **Q1-3. In the following, choose the correct option and give justification.** **(1) From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is:** (A) 7 cm (B) 12 cm (C) 15 cm (D) 24.5 cm **(2) In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that $\angle POQ = 110^\circ$, then $\angle PTQ$ is equal to:** (A) $60^\circ$ (B) $70^\circ$ (C) $80^\circ$ (D) $90^\circ$ **(3) If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of $80^\circ$, then $\angle POA$ is equal to:** (A) $50^\circ$ (B) $60^\circ$ (C) $70^\circ$ (D) $80^\circ$ #### **A1-3. Solution:** **(1) Solution:** Let $OP$ be the radius. $OP \perp PQ$. In $\Delta OPQ$: $OP^2 + PQ^2 = OQ^2 \Rightarrow OP^2 + 24^2 = 25^2 \Rightarrow OP^2 = 625 - 576 = 49$. $OP = 7$ cm. **(Option A)** **(2) Solution:** $OPTQ$ is a quadrilateral. $\angle P = \angle Q = 90^\circ$. $\angle P + \angle O + \angle Q + \angle T = 360^\circ \Rightarrow 90^\circ + 110^\circ + 90^\circ + \angle T = 360^\circ \Rightarrow \angle T = 70^\circ$. **(Option B)** **(3) Solution:** In $\Delta OPA$ and $\Delta OPB$, $\angle OPA = \angle OPB = 80^\circ/2 = 40^\circ$. In $\Delta OPA$, $\angle POA = 180^\circ - (90^\circ + 40^\circ) = 50^\circ$. **(Option A)** #### **Q4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.** #### **A4. Solution:** Let $AB$ be the diameter and $l, m$ be tangents at $A$ and $B$. Radius $OA \perp l$ and $OB \perp m \Rightarrow \angle OAX = 90^\circ$ and $\angle OBY = 90^\circ$. These are alternate interior angles. Since they are equal, lines $l$ and $m$ must be parallel. #### **Q5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.** #### **A5. Solution:** Let $AB$ be the tangent at $P$. Suppose the perpendicular at $P$ does not pass through centre $O$, but through $O'$. Then $\angle O'PB = 90^\circ$. But we know $\angle OPB = 90^\circ$ (radius $\perp$ tangent). This implies $\angle O'PB = \angle OPB$, which is only possible if $O'$ and $O$ coincide. Thus, the perpendicular passes through the centre. #### **Q6. The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.** #### **A6. Solution:** Let $r$ be the radius. Distance from centre = 5 cm, Tangent length = 4 cm. By Pythagoras theorem: $r^2 + 4^2 = 5^2$ $r^2 = 25 - 16 = 9 \Rightarrow r = 3$ cm. The radius is 3 cm. #### **Q7. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.** #### **A7. Solution:** The chord of the larger circle is a tangent to the smaller circle. Let $AB$ be the chord. Radius of smaller circle ($OP$) = 3 cm, Radius of larger circle ($OA$) = 5 cm. $OP \perp AB$ and bisects it. In $\Delta OPA$: $AP^2 = 5^2 - 3^2 = 16 \Rightarrow AP = 4$ cm. Length of chord $AB = 2 \times AP = 8$ cm. #### **Q8. A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC.** #### **A8. Solution:** Tangents from an external point are equal. $AP = AS$, $BP = BQ$, $CR = CQ$, $DR = DS$. Adding LHS and RHS: $(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)$ $AB + CD = AD + BC$. (Proved) #### **Q9. In Fig. 10.13, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B. Prove that $\angle AOB = 90^\circ$.** #### **A9. Solution:** Join $OC$. In $\Delta OPA$ and $\Delta OCA$, $OP=OC, OA=OA, AP=AC \Rightarrow \Delta OPA \cong \Delta OCA$. So, $\angle POA = \angle COA$ (say $x$). Similarly, $\Delta OQB \cong \Delta OCB \Rightarrow \angle QOB = \angle COB$ (say $y$). $PQ$ is a diameter, so $2x + 2y = 180^\circ \Rightarrow x + y = 90^\circ$. $\angle AOB = \angle COA + \angle COB = x + y = 90^\circ$. #### **Q10. Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.** #### **A10. Solution:** Let $PA$ and $PB$ be tangents. $\angle OAP = \angle OBP = 90^\circ$. In quad $OAPB$, $\angle AOB + \angle OAP + \angle APB + \angle OBP = 360^\circ$ $\angle AOB + 90^\circ + \angle APB + 90^\circ = 360^\circ$ $\angle AOB + \angle APB = 180^\circ$. (Supplementary) #### **Q11. Prove that the parallelogram circumscribing a circle is a rhombus.** #### **A11. Solution:** Let $ABCD$ be the parallelogram. From Q8, $AB + CD = AD + BC$. In a parallelogram, $AB = CD$ and $AD = BC$. $2AB = 2AD \Rightarrow AB = AD$. Since adjacent sides are equal, $ABCD$ is a rhombus. #### **Q12. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC.** #### **A12. Solution:** Let $AF = AE = x$. Then sides are $14, x+6, x+8$. Semi-perimeter $s = 14+x$. Area using Heron's $= \sqrt{(14+x)(x)(8)(6)}$. Area also $= \text{Area}(\Delta OBC + \Delta OCA + \Delta OAB) = \frac{1}{2} \times 4 \times (14 + x+6 + x+8) = 2(28+2x) = 4(14+x)$. Equating: $16(14+x)^2 = 48x(14+x) \Rightarrow 14+x = 3x \Rightarrow 2x = 14 \Rightarrow x = 7$. Sides: $AC = 7+6=13$ cm, $AB = 7+8=15$ cm. #### **Q13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.** #### **A13. Solution:** Join the vertices and points of contact to the centre. Eight triangles are formed. By congruency, the 8 angles at the centre are equal in pairs: $\angle 1=\angle 2, \angle 3=\angle 4, \angle 5=\angle 6, \angle 7=\angle 8$. $2(\angle 1 + \angle 8 + \angle 4 + \angle 5) = 360^\circ$ $(\angle 1 + \angle 8) + (\angle 4 + \angle 5) = 180^\circ \Rightarrow \angle AOB + \angle COD = 180^\circ$.