Circles — CBSE Class 10 Maths Textbook Solutions

A circle has infinitely many points lying on it, and by Theorem 10.1 there is exactly one tangent at each point of the circle.

Since a circle has infinitely many points on its boundary, and each such point gives a distinct tangent line, a circle has infinitely many tangents — one at every point on it.

(i) one — this is the defining property of a tangent: it touches the circle at exactly one point.

(ii) secant — a line that cuts a circle in two distinct points is called a secant.

(iii) two — tangents can be drawn parallel to a given line (or to each other) only at the two ends of the diameter perpendicular to that direction; no more than two parallel tangents are possible on one circle.

(iv) point of contact — the single common point of a tangent and the circle is called its point of contact.

By Theorem 10.1, the tangent $PQ$PQ at the point of contact $P$P is perpendicular to the radius $OP$OP. So $\triangle OPQ$OPQ is right-angled at $P$P, with $OQ$OQ as the hypotenuse.

By the Pythagoras theorem:

$OQ^2 = OP^2 + PQ^2$OQ^2 = OP^2 + PQ^2
$12^2 = 5^2 + PQ^2$12^2 = 5^2 + PQ^2
$144 = 25 + PQ^2$144 = 25 + PQ^2
$PQ^2 = 119$PQ^2 = 119
$PQ = \sqrt{119}\text{ cm}$PQ = √119 cm

Answer: (D) $\sqrt{119}$√119 cm.

Construction steps:

1. Draw any line $l$l.
2. Draw a circle with a convenient centre $O$O and radius $r$r, drawn anywhere on the page (not touching $l$l).
3. Draw a line through the topmost point of the circle, parallel to $l$l. This line meets the circle in exactly one point, so it is a tangent to the circle, and it is parallel to $l$l by construction.
4. Draw a second line, also parallel to $l$l, but placed so that it passes through the interior of the circle (between the tangent line and the centre $O$O). This line cuts the circle in two points, so it is a secant, and it too is parallel to $l$l.

This construction shows that a circle can have a tangent and a secant that are both parallel to the same given line $l$l — the tangent just grazes the boundary of the circle while the secant passes through its interior.

Let $O$O be the centre and $P$P the point of contact, so $OQ = 25$OQ = 25 cm and $QP = 24$QP = 24 cm (the tangent length).

Since the tangent is perpendicular to the radius at the point of contact, $\triangle OPQ$OPQ is right-angled at $P$P:

$OQ^2 = OP^2 + QP^2$OQ^2 = OP^2 + QP^2
$25^2 = OP^2 + 24^2$25^2 = OP^2 + 24^2
$625 = OP^2 + 576$625 = OP^2 + 576
$OP^2 = 49 \Rightarrow OP = 7\text{ cm}$OP^2 = 49 OP = 7 cm

Answer: (A) 7 cm.

In quadrilateral $OPTQ$OPTQ, the radii meet the tangents at right angles (Theorem 10.1), so $\angle OPT = \angle OQT = 90^\circ$OPT = OQT = 90^.

The angle sum of a quadrilateral is $360^\circ$360^:

$\angle POQ + \angle OPT + \angle PTQ + \angle OQT = 360^\circ$POQ + OPT + PTQ + OQT = 360^
$110^\circ + 90^\circ + \angle PTQ + 90^\circ = 360^\circ$110^ + 90^ + PTQ + 90^ = 360^
$\angle PTQ = 360^\circ - 290^\circ = 70^\circ$PTQ = 360^ - 290^ = 70^

Answer: (B) $70^\circ$70^.

In right triangle $OAP$OAP (right-angled at $A$A, since $OA \perp AP$OA AP), the line $OP$OP bisects $\angle APB$APB (Theorem 10.2, Remark 2), so

$\angle OPA = \frac{80^\circ}{2} = 40^\circ$OPA = 80^/2 = 40^

The angles of $\triangle OAP$OAP sum to $180^\circ$180^:

$\angle OAP + \angle OPA + \angle AOP = 180^\circ$OAP + OPA + AOP = 180^
$90^\circ + 40^\circ + \angle AOP = 180^\circ$90^ + 40^ + AOP = 180^
$\angle AOP = 50^\circ$AOP = 50^

So $\angle POA = 50^\circ$POA = 50^.

Answer: (A) $50^\circ$50^.

Given: A circle with centre $O$O and diameter $AB$AB. Let $l$l be the tangent at $A$A and $m$m be the tangent at $B$B.

To prove: $l \parallel m$l m.

Proof:

1. Since $l$l is the tangent at $A$A, by Theorem 10.1, $OA \perp l$OA l, i.e., $\angle OAl = 90^\circ$OAl = 90^.
2. Since $m$m is the tangent at $B$B, by Theorem 10.1, $OB \perp m$OB m, i.e., $\angle OBm = 90^\circ$OBm = 90^.
3. Now $A$A, $O$O, $B$B all lie on the single straight line $AB$AB (since $AB$AB is a diameter).
4. So $l$l and $m$m are both perpendicular to the same straight line $AB$AB, at the two points $A$A and $B$B on it.
5. Two lines that are perpendicular to the same line are parallel to each other.

Therefore $l \parallel m$l m. Hence proved.

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

To prove: The line perpendicular to $XY$XY at $P$P passes through $O$O.

Proof:

1. By Theorem 10.1, the radius $OP$OP is perpendicular to the tangent $XY$XY at the point of contact $P$P. So $OP$OP is a line through $P$P that is perpendicular to $XY$XY.
2. Through a given point on a line, exactly one perpendicular can be drawn to that line (in a plane).
3. Since $OP$OP is already such a perpendicular at $P$P, any other line drawn perpendicular to $XY$XY at $P$P must coincide with $OP$OP — there cannot be a second, different perpendicular at the same point.
4. As the line $OP$OP passes through the centre $O$O, it follows that the perpendicular to the tangent at the point of contact necessarily passes through the centre of the circle.

Hence proved.

Let $O$O be the centre and $P$P the point of contact, so $OA = 5$OA = 5 cm and $AP = 4$AP = 4 cm (the tangent length).

Since the tangent is perpendicular to the radius at the point of contact (Theorem 10.1), $\triangle OPA$OPA is right-angled at $P$P:

$OA^2 = OP^2 + AP^2$OA^2 = OP^2 + AP^2
$5^2 = OP^2 + 4^2$5^2 = OP^2 + 4^2
$25 = OP^2 + 16$25 = OP^2 + 16
$OP^2 = 9 \Rightarrow OP = 3\text{ cm}$OP^2 = 9 OP = 3 cm

The radius of the circle is 3 cm.

Let $O$O be the common centre, with $OA = 5$OA = 5 cm the radius of the larger circle and $OP = 3$OP = 3 cm the radius of the smaller circle, where the chord $AB$AB of the larger circle touches the smaller circle at $P$P.

Since $AB$AB is tangent to the smaller circle at $P$P, $OP \perp AB$OP AB (Theorem 10.1).

A perpendicular drawn from the centre of a circle to a chord bisects the chord, so $P$P is the midpoint of $AB$AB, i.e., $AP = PB$AP = PB.

In right triangle $OPA$OPA (right-angled at $P$P):

$OA^2 = OP^2 + AP^2$OA^2 = OP^2 + AP^2
$5^2 = 3^2 + AP^2$5^2 = 3^2 + AP^2
$25 = 9 + AP^2$25 = 9 + AP^2
$AP^2 = 16 \Rightarrow AP = 4\text{ cm}$AP^2 = 16 AP = 4 cm

So $AB = 2 \times AP = 2 \times 4 = 8$AB = 2 × AP = 2 × 4 = 8 cm.

The length of the chord is 8 cm.

Given: A quadrilateral $ABCD$ABCD circumscribing a circle, i.e., each of its four sides touches the circle.

To prove: $AB + CD = AD + BC$AB + CD = AD + BC.

Proof: Let the circle touch the sides $AB$AB, $BC$BC, $CD$CD, $DA$DA at points $P$P, $Q$Q, $R$R, $S$S respectively.

Since the lengths of tangents drawn from an external point to a circle are equal (Theorem 10.2):

• From vertex $A$A: $AP = AS$AP = AS
• From vertex $B$B: $BP = BQ$BP = BQ
• From vertex $C$C: $CR = CQ$CR = CQ
• From vertex $D$D: $DR = DS$DR = DS

Now add the left-hand side pairs:

$AB + CD = (AP + PB) + (CR + RD)$AB + CD = (AP + PB) + (CR + RD)

Substituting the equal tangent lengths:

$AB + CD = (AS + BQ) + (CQ + DS)$AB + CD = (AS + BQ) + (CQ + DS)

Rearranging the terms on the right:

$AB + CD = (AS + DS) + (BQ + CQ) = AD + BC$AB + CD = (AS + DS) + (BQ + CQ) = AD + BC

Therefore $AB + CD = AD + BC$AB + CD = AD + BC. Hence proved.

Given: $XY$XY and $X'Y'$X'Y' are parallel tangents to a circle with centre $O$O, touching it at $P$P and $Q$Q respectively. A third tangent $AB$AB touches the circle at $C$C, and meets $XY$XY at $A$A and $X'Y'$X'Y' at $B$B.

To prove: $\angle AOB = 90^\circ$AOB = 90^.

Proof: Join $OA$OA, $OB$OB and $OC$OC.

From point $A$A, the two tangents are $AP$AP (along $XY$XY) and $AC$AC (along $AB$AB). Since $OA$OA joins $A$A to the centre, it bisects the angle between these two tangents (Theorem 10.2, Remark 2):

$\angle PAC = 2\,\angle OAC \quad \text{i.e.} \quad \angle OAC = \angle OAP$PAC = 2\, OAC i.e. OAC = OAP

Similarly, from point $B$B, the tangents are $BQ$BQ (along $X'Y'$X'Y') and $BC$BC (along $AB$AB), and $OB$OB bisects $\angle QBC$QBC:

$\angle OBC = \angle OBQ$OBC = OBQ

Since $XY \parallel X'Y'$XY X'Y' and $AB$AB is a transversal cutting them, the co-interior (same-side interior) angles are supplementary:

$\angle PAB + \angle QBA = 180^\circ$PAB + QBA = 180^

i.e. $\angle PAC + \angle QBC = 180^\circ$PAC + QBC = 180^ (as $C$C lies on $AB$AB).

Since $OA$OA and $OB$OB bisect these angles respectively:

$2\,\angle OAC + 2\,\angle OBC = 180^\circ \Rightarrow \angle OAC + \angle OBC = 90^\circ$2\, OAC + 2\, OBC = 180^ OAC + OBC = 90^

Now, in $\triangle OAB$OAB, the angles sum to $180^\circ$180^:

$\angle OAB + \angle OBA + \angle AOB = 180^\circ$OAB + OBA + AOB = 180^
$\angle OAC + \angle OBC + \angle AOB = 180^\circ$OAC + OBC + AOB = 180^
$90^\circ + \angle AOB = 180^\circ$90^ + AOB = 180^
$\angle AOB = 90^\circ$AOB = 90^

Hence proved.

Given: A circle with centre $O$O, and two tangents $PA$PA, $PB$PB drawn from an external point $P$P, touching the circle at $A$A and $B$B.

To prove: $\angle APB + \angle AOB = 180^\circ$APB + AOB = 180^.

Proof: Join $OA$OA and $OB$OB.

Since a tangent to a circle is perpendicular to the radius through the point of contact (Theorem 10.1):

$\angle OAP = 90^\circ \qquad \text{and} \qquad \angle OBP = 90^\circ$OAP = 90^ and OBP = 90^

Now consider the quadrilateral $OAPB$OAPB. The sum of its interior angles is $360^\circ$360^:

$\angle OAP + \angle APB + \angle PBO + \angle BOA = 360^\circ$OAP + APB + PBO + BOA = 360^

Substituting the two right angles:

$90^\circ + \angle APB + 90^\circ + \angle AOB = 360^\circ$90^ + APB + 90^ + AOB = 360^
$\angle APB + \angle AOB = 360^\circ - 180^\circ = 180^\circ$APB + AOB = 360^ - 180^ = 180^

So the angle between the two tangents ($\angle APB$APB) is supplementary to the angle subtended at the centre by the segment joining the points of contact ($\angle AOB$AOB). Hence proved.

Given: A parallelogram $ABCD$ABCD circumscribing a circle, touching $AB$AB, $BC$BC, $CD$CD, $DA$DA at $P$P, $Q$Q, $R$R, $S$S respectively.

To prove: $ABCD$ABCD is a rhombus.

Proof: By Theorem 10.2, tangent lengths from each vertex are equal:

$AP = AS, \quad BP = BQ, \quad CR = CQ, \quad DR = DS$AP = AS, BP = BQ, CR = CQ, DR = DS

Adding all four equations:

$AP + BP + CR + DR = AS + BQ + CQ + DS$AP + BP + CR + DR = AS + BQ + CQ + DS
$(AP+PB) + (CR+RD) = (AS+SD) + (BQ+QC)$(AP+PB) + (CR+RD) = (AS+SD) + (BQ+QC)
$AB + CD = AD + BC \qquad \dots (1)$AB + CD = AD + BC (1)

But $ABCD$ABCD is a parallelogram, so its opposite sides are equal:

$AB = CD \qquad \text{and} \qquad AD = BC \qquad \dots (2)$AB = CD and AD = BC (2)

Substituting (2) into (1):

$AB + AB = BC + BC \Rightarrow 2AB = 2BC \Rightarrow AB = BC$AB + AB = BC + BC 2AB = 2BC AB = BC

So two adjacent sides $AB$AB and $BC$BC are equal. Combined with $AB = CD$AB = CD and $AD = BC$AD = BC from (2), all four sides are equal:

$AB = BC = CD = DA$AB = BC = CD = DA

A parallelogram with all four sides equal is, by definition, a rhombus. Hence proved.

Let the incircle touch $BC$BC, $CA$CA, $AB$AB at $D$D, $E$E, $F$F respectively. Using equal tangent lengths from each vertex (Theorem 10.2):

$BD = BF = 8\text{ cm}, \qquad CD = CE = 6\text{ cm}, \qquad AE = AF = x \text{ (say)}$BD = BF = 8 cm, CD = CE = 6 cm, AE = AF = x (say)

Then:

$AB = AF + FB = x + 8, \qquad AC = AE + EC = x + 6, \qquad BC = BD + DC = 14\text{ cm}$AB = AF + FB = x + 8, AC = AE + EC = x + 6, BC = BD + DC = 14 cm

The semi-perimeter of $\triangle ABC$ABC:

$s = \frac{AB+BC+CA}{2} = \frac{(x+8)+14+(x+6)}{2} = x + 14$s = AB+BC+CA/2 = (x+8)+14+(x+6)/2 = x + 14

so $s - BC = x$s - BC = x, $s - AC = 8$s - AC = 8, $s - AB = 6$s - AB = 6.

By Heron's formula:

$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(x+14)\cdot x \cdot 6 \cdot 8} = \sqrt{48x(x+14)}$Area = √s(s-a)(s-b)(s-c) = √(x+14)· x · 6 · 8 = √48x(x+14)

Also, using the incircle radius $r = 4$r = 4 cm and Area $= r \times s$= r × s:

$\text{Area} = 4(x+14)$Area = 4(x+14)

Equating the two expressions for area and squaring both sides:

$[4(x+14)]^2 = 48x(x+14)$[4(x+14)]^2 = 48x(x+14)
$16(x+14)^2 = 48x(x+14)$16(x+14)^2 = 48x(x+14)

Dividing both sides by $(x+14)$(x+14) (which is non-zero):

$16(x+14) = 48x$16(x+14) = 48x
$16x + 224 = 48x$16x + 224 = 48x
$224 = 32x \Rightarrow x = 7$224 = 32x x = 7

Therefore:

$AB = x + 8 = 15\text{ cm} \qquad \text{and} \qquad AC = x + 6 = 13\text{ cm}$AB = x + 8 = 15 cm and AC = x + 6 = 13 cm

Given: A quadrilateral $ABCD$ABCD circumscribes a circle with centre $O$O, touching $AB$AB, $BC$BC, $CD$CD, $DA$DA at $P$P, $Q$Q, $R$R, $S$S respectively.

To prove: $\angle AOB + \angle COD = 180^\circ$AOB + COD = 180^ and $\angle BOC + \angle AOD = 180^\circ$BOC + AOD = 180^.

Proof: Join $OA$OA, $OB$OB, $OC$OC, $OD$OD and also $OP$OP, $OQ$OQ, $OR$OR, $OS$OS.

In $\triangle OAP$OAP and $\triangle OAS$OAS: $OA$OA is common, $OP = OS$OP = OS (radii), and $AP = AS$AP = AS (equal tangents from $A$A). So $\triangle OAP \cong \triangle OAS$OAP OAS (SSS), which gives

$\angle AOP = \angle AOS = a \text{ (say)}$AOP = AOS = a (say)

By the same argument at each vertex:

$\angle BOP = \angle BOQ = b, \qquad \angle COQ = \angle COR = c, \qquad \angle DOR = \angle DOS = d$BOP = BOQ = b, COQ = COR = c, DOR = DOS = d

The eight angles $\angle AOP, \angle BOP, \angle BOQ, \angle COQ, \angle COR, \angle DOR, \angle DOS, \angle AOS$AOP, BOP, BOQ, COQ, COR, DOR, DOS, AOS together make one full turn at $O$O:

$2a + 2b + 2c + 2d = 360^\circ \Rightarrow a+b+c+d = 180^\circ \qquad \dots (1)$2a + 2b + 2c + 2d = 360^ a+b+c+d = 180^ (1)

Now, $\angle AOB = \angle AOP + \angle POB = a + b$AOB = AOP + POB = a + b, and $\angle COD = \angle COR + \angle ROD = c + d$COD = COR + ROD = c + d. Adding these:

$\angle AOB + \angle COD = (a+b) + (c+d) = 180^\circ \quad \text{[by (1)]}$AOB + COD = (a+b) + (c+d) = 180^ [by (1)]

Similarly, $\angle BOC = \angle BOQ + \angle QOC = b+c$BOC = BOQ + QOC = b+c and $\angle AOD = \angle AOS + \angle SOD = a+d$AOD = AOS + SOD = a+d, so

$\angle BOC + \angle AOD = (b+c) + (a+d) = 180^\circ \quad \text{[by (1)]}$BOC + AOD = (b+c) + (a+d) = 180^ [by (1)]

So the pair of opposite sides $AB, CD$AB, CD subtends angles at $O$O that are supplementary, and likewise the pair $BC, AD$BC, AD. Hence proved.