NCERT Solutions For Class 10 Maths: Circles

EXERCISE 10.1

Question 1: How many tangents can a circle have?

Solution-
1.A circle can have infinitely many tangents.

2.This is because there is exactly one tangent corresponding to every point on the circle, and a circle consists of infinitely many points.

Question 2: Fill in the blanks :
(i) A tangent to a circle intersects it in $\underline{\hspace{1cm}}$ point (s).
(ii) A line intersecting a circle in two points is called a $\underline{\hspace{1cm}}$.
(iii) A circle can have $\underline{\hspace{1cm}}$ parallel tangents at the most.
(iv) The common point of a tangent to a circle and the circle is called $\underline{\hspace{1cm}}$.

Solution-
(i) A tangent to a circle intersects it in one point (s).

(ii) A line intersecting a circle in two points is called a secant.

(iii) A circle can have two parallel tangents at the most.

(iv) The common point of a tangent to a circle and the circle is called the point of contact.

Question 3: A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is :
(A) 12 cm (B) 13 cm (C) 8.5 cm (D) $\sqrt{119}$ cm.

Solution-
1.By Theorem 10.1, the radius OP is perpendicular to the tangent PQ at the point of contact P. Thus, $\triangle OPQ$ is a right-angled triangle at P.

2.We use the Pythagoras Theorem: $OP^2 + PQ^2 = OQ^2$.

3.Given $OP = 5$ cm and $OQ = 12$ cm:
$5^2 + PQ^2 = 12^2$
$25 + PQ^2 = 144$
$PQ^2 = 144 - 25 = 119$

4.$PQ = \sqrt{119}$ cm.

5.The correct option is (D) $\sqrt{119}$ cm.

Question 4: Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.

Solution-
1.Draw a circle with a fixed center O.

2.Draw a line PQ (the given line).

3.Draw a secant (a line intersecting the circle at two points) parallel to the given line PQ.

4.Draw a tangent (a line touching the circle at only one point—the point of contact) parallel to the given line PQ. A circle can have at most two tangents parallel to a given secant.

EXERCISE 10.2

Question 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

Solution-
1.Let P be the point of contact and O be the centre. Since the radius is perpendicular to the tangent (Theorem 10.1), $\triangle OPQ$ is a right triangle at P.

2.Given $PQ = 24$ cm and $OQ = 25$ cm. We find the radius $OP$ using the Pythagoras Theorem:
$OP^2 = OQ^2 - PQ^2$.
$OP^2 = 25^2 - 24^2 = 625 - 576 = 49$.

3.$OP = 7$ cm.

4.The radius of the circle is 7 cm. The correct option is (A) 7 cm.

Question 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) $90^\circ$

Solution-
1.OPTQ is a quadrilateral. By Theorem 10.1, the radii OP and OQ are perpendicular to the tangents TP and TQ respectively.
$\angle OPT = 90^\circ$ and $\angle OQT = 90^\circ$.

2.The sum of the angles in a quadrilateral is $360^\circ$:
$\angle PTQ + \angle POQ + \angle OPT + \angle OQT = 360^\circ$.
$\angle PTQ + 110^\circ + 90^\circ + 90^\circ = 360^\circ$.

3.$\angle PTQ = 360^\circ - 290^\circ = 70^\circ$.

4.The correct option is (B) $70^\circ$.

Question 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) $80^\circ$

Solution-
1.In the quadrilateral OAPB, $\angle OAP = 90^\circ$ and $\angle OBP = 90^\circ$ (Theorem 10.1). Given $\angle APB = 80^\circ$.

2.Consider $\triangle OAP$ and $\triangle OBP$. $OA = OB$ (Radii), $OP = OP$ (Common), and $PA = PB$ (Tangents from an external point P are equal, Theorem 10.2).
$\triangle OAP \cong \triangle OBP$ (RHS congruence).

3.By CPCT, $\angle POA = \angle POB$. Also, $OP$ is the angle bisector of $\angle APB$.
$\angle APO = \frac{1}{2} \angle APB = \frac{1}{2} (80^\circ) = 40^\circ$.

4.In the right triangle $\triangle OAP$:
$\angle POA + \angle OAP + \angle APO = 180^\circ$.
$\angle POA + 90^\circ + 40^\circ = 180^\circ$.

5.$\angle POA = 180^\circ - 130^\circ = 50^\circ$.

6.The correct option is (A) $50^\circ$.

Question 4: Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Solution-
1.Let AB be a diameter of a circle with center O. Let XY and X'Y' be the tangents at the endpoints A and B respectively.

2.Since XY is a tangent at A, by Theorem 10.1, $OA \perp XY$. This means $\angle OAX = 90^\circ$ and $\angle OAY = 90^\circ$.

3.Since X'Y' is a tangent at B, $OB \perp X'Y'$. This means $\angle OBX' = 90^\circ$ and $\angle OBY' = 90^\circ$.

4.Consider the transversal AB intersecting lines XY and X'Y'. The angles $\angle YAB$ (or $\angle OAY$) and $\angle ABX'$ (or $\angle OBX'$) are alternate interior angles.

5.Since $\angle OAY = 90^\circ$ and $\angle OBX' = 90^\circ$, the alternate interior angles are equal.

6.Therefore, the tangents XY and X'Y' are parallel.

Question 5: Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Solution-
1.Let XY be a tangent to a circle with centre O, and P be the point of contact.

2.By Theorem 10.1, the radius OP is perpendicular to the tangent XY at P.

3.We know that for any given line (XY) and a point on it (P), there is one and only one line perpendicular to the given line passing through that point.

4.Since OP is the unique line segment from O that is perpendicular to XY at P, any line drawn perpendicular to the tangent XY at the point of contact P must coincide with the line segment OP, and hence must pass through the centre O.

Question 6: 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.

Solution-
1.Let O be the centre and P be the point of contact. $AP = 4$ cm is the length of the tangent, and $OA = 5$ cm is the distance from the centre to the external point A.

2.By Theorem 10.1, $\triangle OPA$ is a right triangle at P.

3.Using the Pythagoras Theorem: $OP^2 = OA^2 - AP^2$.
$OP^2 = 5^2 - 4^2 = 25 - 16 = 9$.

4.$OP = 3$ cm.

5.The radius of the circle is 3 cm.

Question 7: 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.

Solution-
1.Let O be the common centre. Let $R=5$ cm (radius of larger circle) and $r=3$ cm (radius of smaller circle).

2.Let AB be the chord of the larger circle that touches the smaller circle at P. Thus, AB is a tangent to the smaller circle at the point of contact P. $OP = 3$ cm.

3.By Theorem 10.1, $OP \perp AB$. OA is the radius of the larger circle, $OA = 5$ cm.

4.In the right triangle $\triangle OPA$, using the Pythagoras Theorem:
$AP^2 = OA^2 - OP^2 = 5^2 - 3^2 = 25 - 9 = 16$.
$AP = 4$ cm.

5.Since the perpendicular from the centre bisects the chord (a property of the chord of the larger circle), $AB = 2 \times AP$.
$AB = 2 \times 4 = 8$ cm.

6.The length of the chord is 8 cm.

Question 8: A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that $AB + CD = AD + BC$.

Solution-
1.Let the circle touch the sides AB, BC, CD, and DA at points P, Q, R, and S respectively.

2.By Theorem 10.2, the lengths of tangents drawn from an external point to a circle are equal.
$AP = AS$ (From A)
$BP = BQ$ (From B)
$CR = CQ$ (From C)
$DR = DS$ (From D)

3.Adding the segments on the left side:
$AB + CD = (AP + BP) + (CR + DR)$.

4.Substituting the equal segments from the right side:
$AB + CD = (AS + BQ) + (CQ + DS)$.

5.Rearrange the terms to form the opposite sides of the quadrilateral:
$AB + CD = (AS + DS) + (BQ + CQ)$.

6.Since $AS + DS = AD$ and $BQ + CQ = BC$:
$AB + CD = AD + BC$. Hence proved.

Question 9: 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 $\angle AOB = 90^\circ$.

Solution-
1.We are given two parallel tangents XY and X'Y' and a third tangent AB.

2.Join OA, OB, OC. In triangles $\triangle OAP$ and $\triangle OAC$: $OP = OC$ (Radii), $OA = OA$ (Common), and $AP = AC$ (Tangents from A are equal).
$\triangle OAP \cong \triangle OAC$ (SSS congruence).
By CPCT, $\angle PAO = \angle CAO$. Let $\angle CAO = x$.

3.Similarly, in triangles $\triangle OBQ$ and $\triangle OBC$: $OQ = OC$, $OB = OB$, and $BQ = BC$ (Tangents from B are equal).
$\triangle OBQ \cong \triangle OBC$ (SSS congruence).
By CPCT, $\angle QBO = \angle CBO$. Let $\angle CBO = y$.

4.Since XY is parallel to X'Y', the line segment PQ (a diameter passing through O) acts as a transversal, and the angles on the same side of the transversal are supplementary: $\angle PAB + \angle QBA = 180^\circ$.

5.Since $\angle PAB = 2x$ and $\angle QBA = 2y$:
$2x + 2y = 180^\circ$.
$x + y = 90^\circ$.

6.In $\triangle AOB$, the sum of angles is $180^\circ$:
$\angle AOB + \angle OAB + \angle OBA = 180^\circ$.
$\angle AOB + x + y = 180^\circ$.

7.Substitute $x + y = 90^\circ$:
$\angle AOB + 90^\circ = 180^\circ$.
$\angle AOB = 90^\circ$. Hence proved.

Question 10: 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.

Solution-
1.Let T be the external point, and TP and TQ be the tangents with points of contact P and Q. O is the centre. We need to prove $\angle PTQ + \angle POQ = 180^\circ$.

2.Consider the quadrilateral OPTQ.

3.By Theorem 10.1, the radius is perpendicular to the tangent at the point of contact.
$\angle OPT = 90^\circ$.
$\angle OQT = 90^\circ$.

4.The sum of the internal angles of the quadrilateral OPTQ is $360^\circ$:
$\angle PTQ + \angle POQ + \angle OPT + \angle OQT = 360^\circ$.
$\angle PTQ + \angle POQ + 90^\circ + 90^\circ = 360^\circ$.

5.$\angle PTQ + \angle POQ = 360^\circ - 180^\circ = 180^\circ$.

6.Since the sum is $180^\circ$, the angle between the two tangents ($\angle PTQ$) and the angle subtended by the line-segment joining the points of contact at the centre ($\angle POQ$) are supplementary. Hence proved.

Question 11: Prove that the parallelogram circumscribing a circle is a rhombus.

Solution-
1.Let ABCD be a parallelogram circumscribing a circle. By the definition of a parallelogram, opposite sides are equal: $AB = CD$ and $AD = BC$.

2.Since the quadrilateral circumscribes the circle, according to the result proved in Question 8:
$AB + CD = AD + BC$.

3.Substitute $CD = AB$ and $BC = AD$ into the equation:
$AB + AB = AD + AD$.
$2AB = 2AD$.
$AB = AD$.

4.Since adjacent sides $AB$ and $AD$ are equal, and the figure is already a parallelogram, all four sides must be equal ($AB = BC = CD = DA$).

5.A parallelogram with all sides equal is a rhombus. Hence proved.

Question 12: 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 (see Fig. 10.14). Find the sides AB and AC.

Solution-
1.Let the circle with center O and radius $r=4$ cm touch sides AB and AC at F and E respectively.

2.Using Theorem 10.2 (lengths of tangents from an external point are equal):
$BF = BD = 8$ cm.
$CE = CD = 6$ cm.
Let $AF = AE = x$ cm.

3.The side lengths are: $AB = x + 8$, $AC = x + 6$, $BC = 8 + 6 = 14$.

4.Calculate the semi-perimeter $s$:
$s = \frac{(x + 8) + (x + 6) + 14}{2} = \frac{2x + 28}{2} = x + 14$.

5.The Area of $\triangle ABC$ can be expressed as $Area = r \cdot s$ (where r is the inradius):
$Area = 4(x + 14)$.

6.The Area of $\triangle ABC$ can also be calculated using Heron's formula:
$Area = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(x + 14)(x)(8)(6)} = \sqrt{48x(x + 14)}$.
$(s-a = x, s-b = 8, s-c = 6)$.

7.Equating the two area expressions and squaring:
$4(x + 14) = \sqrt{48x(x + 14)}$.
$16(x + 14)^2 = 48x(x + 14)$.
$16(x + 14) = 48x$.
$x + 14 = 3x$.
$2x = 14$, so $x = 7$ cm.

8.Calculate the sides AB and AC:
$AB = x + 8 = 7 + 8 = 15$ cm.
$AC = x + 6 = 7 + 6 = 13$ cm.

Question 13: Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Solution-
1.Let ABCD be a quadrilateral circumscribing a circle with center O. Let the points of contact be P, Q, R, S on sides AB, BC, CD, DA respectively. We must prove $\angle AOB + \angle COD = 180^\circ$ and $\angle BOC + \angle AOD = 180^\circ$.

2.Join OA, OB, OC, OD, OP, OQ, OR, OS.

3.By Theorem 10.2, tangents from an external point are equal ($AP=AS$, $BP=BQ$, etc.). By Theorem 10.1, the radii are perpendicular to the tangents.

4.Consider $\triangle AOP$ and $\triangle AOS$. They are congruent (RHS or SSS congruence). Let $\angle AOP = \angle AOS = a$.

5.Similarly, let $\angle BOP = \angle BOQ = b$; $\angle COQ = \angle COR = c$; and $\angle DOR = \angle DOS = d$.

6.The sum of all angles around the centre O is $360^\circ$:
$2a + 2b + 2c + 2d = 360^\circ$.
$a + b + c + d = 180^\circ$.

7.The angle subtended by side AB is $\angle AOB = a + b$. The angle subtended by the opposite side CD is $\angle COD = c + d$.
$\angle AOB + \angle COD = (a + b) + (c + d) = 180^\circ$.

8.The angle subtended by side BC is $\angle BOC = b + c$. The angle subtended by the opposite side AD is $\angle AOD = a + d$.
$\angle BOC + \angle AOD = (b + c) + (a + d) = 180^\circ$.

9.Thus, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre. Hence proved.