Solutions-Class-10-Mathematics-2-Chapter-3-Circle-Maharashtra Board

(1) ∠ CAB = 90°     ... (Tangent theorem) ... (1)

(2) seg CA ⊥ line AB

CA = 6 cm      ... (Given)

∴ C is at a distance of 6 cm from line AB.

(3) In Δ CAB,

∠ CAB = 90°       ... [From (1)]

∴ by Pythagoras theorem,

CB² = CA² + AB²

∴ BC² = 6² + 6²

∴ BC² = 36 + 36 = 72

∴ BC = 6\(\sqrt{2}\) cm                ... (Taking square roots of both the sides)

d(B,C) = BC = 6\(\sqrt{2}\) cm.

(4) In Δ ABC,

CA = AB                               ... (Each is the length 6 cm)

∠ ABC = ∠ ACB                     ... (Isosceles triangle theorem)

Let ∠ ABC = ∠ ACB = x°

In Δ ABC,

∠ CAB + ∠ ABC + ∠ ACB = 180°      ... (Sum of all angles of a triangle is 180°)

∴ 90° + x° + x° = 180°

∴ 2x° = 180°— 90°

∴  2x° = 90°

∴  x° = 90°/2

∴ x° = 45°

∴ ∠ ABC = 45°.

Answer :

(1) ∠ CAB= 90°,

(2) C is at a distance of 6 cm from line AB,

(3) d(B, C) = 6\(\sqrt{2}\) cm and

(4) ∠ ABC = 45°.

Draw seg OM and seg ON.

(1) In Δ OMR,

∠ OMR = 90°     ... (Tangent theorem)

∴ by Pythagoras theorem,

OR² = OM² + MR²

∴ 10² = 5² + MR²

∴ MR² = 100 — 25

∴ MR² = 75

∴ MR = 5\(\sqrt{3}\) cm        ... (Taking square roots of both the sides)

RM = RN = 5\(\sqrt{3}\) cm        ... (Tangent segment theorem)

Length of each tangent segment is 5\(\sqrt{3}\) cm.    ... (Given)

(2) In Δ RMO, ∠ OMR = 90°   ... (Tangent theorem)

OM = 5 cm and OR = 10 cm

∴ OM = \(\frac{1}{2}\)OR

∴ ∠ MRO = 30°   ... (Converse of 30°—60°—90° triangle theorem)

Similarly ∠ NRO = 30°

(3) ∠ MRN = ∠ MRO + ∠ NRO   ... (Angle addition postulate)

∴ ∠ MRN = 30°+ 30°

∴ ∠ MRN= 60°

Answer :

(1) Length of each tangent segment is 5\(\sqrt{3}\) cm,

(2) ∠ MRO = 30°

(3) ∠ MRN = 60°.

Proof : In Δ RMO and Δ RNO

∠ RMO = ∠ RNO = 90°   ... (Tangent theorem)

Hypotenuse OR ≅ hypotenuse OR         ... (Common side)

Side OM ≅ side ON           ... (Radii of the same circle)

∴ Δ RMO ≅ Δ RNO            ... (Hypotenuse side theorem)

∴ ∠ MRO ≅ ∠ NRO and ∠ MOR ≅ ∠ NOR          ... (c.a.c.t.)

∴ seg OR bisects ∠ MRN and ∠ MON.

Let the centre of the circle be O. Let the tangents AB and CD touch the circle at points M and N respectively.

Draw seg OM and seg ON.

Through O, draw line XY || line AB

Line AB || line CD             ... (Given)

Line XY || line AB               ... (Construction)

∴ line AB || line XY || line CD   ... (1)

∠ OMA = ∠ ONC = 90°               ... (Tangent theorem)

Line AB || line XY and line MO is the transversal,

∠ AMO + ∠ MOX = 180°              ... (Interior angles theorem)

∴ 90°+ ∠ MOX = 180°

∴  ∠ MOX = 180°— 90°

∴  ∠ MOX = 90°

Similarly ∠ NOX = 90°

∴ ∠ MOX + ∠ NOX = 90° + 90° = 180°

∴  ∠ MOX and ∠ NOX are adjacent as well as supplementary.

∴ ∠ MOX and ∠ NOX form linear pair.

∴ ray OM and ray ON are opposite rays.

∴ points M, O and N are collinear.

∴ MN is the distance between the parallel tangents AB and CD

MN = 2 × radius         ... (Diameter is twice the radius)

∴ MN =2 × 4.5

∴ MN = 9 cm

Answer :  Distance between parallel tangents is 9 cm.

Let the radii of the two circles be r₁ and r₂.

r₁ = 3.5 cm and r₂ = 4.8 cm     ... (Given)

The two circles are touching each other internally.

∴ by theorem of touching circles, the centres and the point of contact are collinear.

The distance between the centres of the circles touching internally is equal to the difference of their radii

∴ distance between their centres = r₂ — r₁ = 4.8 — 3.5 = 1.3 cm.

Answer : Distance between the centres is 1.3 cm.

Let the radii of the two circles be r₁ and r₂.

r₁ = 5.5 cm and r₂ = 4.2 cm      ... (Given)

The two circles are touching each other externally

∴ by theorem of touching circles, the centres and the point of contact are collinear.

The distance between the centres of the circles touching externally is equal to the sum of their radii.

∴ distance between their centres = r₁ + r₂ = 5.5 + 4.2 = 9.7 cm

Answer : Distance between the centres is 9.7 cm.

(i) Circles touch each other externally.

(i) Circles touch each other internally.

(1) P—R—Q               ... (By theorem of touching circles)

In Δ PAR,

seg PA ≅ seg PR            ... (Radii of the same circle)

∴ ∠ PAR ≅ ∠ PRA          ... (Isosceles triangle theorem) ... (1)

In Δ QRB,

seg QR ≅ seg QB          ... (Radii of the same circle)

∴ ∠ QRB ≅ ∠ QBR     ... (Isosceles triangle theorem) ... (2)

∠ PRA ≅ ∠ QRB         ... (Vertically opposite angles) ... (3)

∴ from (1). (2) and (3), we get,

∠ PAR ≅ ∠ QBR

i.e. ∠ PAB ≅ ∠ QBA             ... (A—R—B)

∴  seg AP || seg BQ   ... (Alternate angles test for parallel lines)

(2) In Δ APR and Δ RQB,

∠ PAR ≅ ∠ QRB            ... [From (1) and (3)]

∠ PRA ≅ ∠ QBR           ... [From (2) and (3)]

∴ Δ APR ~ Δ RQB        ... (AA test of similarity)

(3) ∠ PAR = 35°           ... (Given)

∴ from (1), (2) and (3),

∠ PRA = ∠ QRB = ∠ QBR = 35°         ... (4)

In Δ QRB,

∠ QRB + ∠ QBR + ∠ RQB = 180°        ... (Sum of all angles of a triangle is 180°)

∴ 35° + 35° + ∠ RQB = 180°              ... [From (4)]

∴ 70° + ∠ RQB = 180°

∴ ∠ RQB = 180°— 70°

∴ ∠ RQB = 110°

∠ RQB = 110°.

Draw seg AC, seg BD and seg AB. Draw seg AM ⊥ seg BD and B—M—D.

A—E—B   ... (By theorem of touching circles)

AE = AC = 4 cm and BE = BD =6 cm  ... (Radii of the same circle)

∠ ACD = ∠ BDC = 90°                ... (Tangent theorem) ... (1)

In □ AMDC,

∠ ACD + ∠ CDM + ∠ AMD + ∠ CAM = 360°   …  (Sum of all angles of quadrilateral is 360°)

∴ 90° + 90° + 90° + ∠ CAM = 360°   [From construction, (1) and B—M—D]

∴ 270° + ∠ CAM = 360°

∴ ∠CAM = 360° — 270°

∴ ∠  CAM = 90°

∴ □ AMDC is a rectangle     ….(By definition)

∴ DM = AC = 4 cm    ….. (Opposite sides of rectangle are equal)

BM + MD = BD      ….(B—M—D)

∴ BM + 4 = 6

∴ BM = 6 — 4

∴ BM = 2 cm.

∴AB =AE + EB  (A—E—B)

∴AB = 4 + 6

∴AB = 10 cm.

In Δ AMB, ∠ AMB = 90°.      …..(Construction)

∴ by Pythagoras theorem,

AB² = AM² + BM²

∴ 10² = AM² + 2²

∴ AM² = 10² — 2²

∴ AM² = 100—4  ∴ AM² = 96 = 16 × 6

∴ AM = 4  cm    …. (Taking square roots of both the sides)

CD = AM     …. (Opposite sides of rectangle are equal)

∴ CD = 4  cm

Answer : CD = 4  cm.

m(arc EF) = ∠ ECF       … (Definition of measure of minor arc)

∠ ECF = 70°     ….(Given)

∴ m(arc EF) = 70°

m(arc DGF) + m(arc EF) + m(arc DE) = 360°     …. (Measure of a circle is 360°)

∴ 200° + 70° + m(arc DE) = 360°

∴ 270° + m(arc DE) = 360°

∴ m(arc DE) = 360° — 270°

∴ m(arc DE) = 90°

m(arc DE) + m(arc EF) = m(arc DEF)     ….. (Arc addition postulate)

∴ 90° + 70° = m(arc DEF)

∴ m(arc DEF) = 160°

Answer : m(arc DE) = 90° and m(arc DEF) = 160°.

Proof : Δ QRS is an equilateral triangle   …..(given)

seg RS ≅ seg QS ≅ seg QR     … (Sides of an equilateral triangle)

Arcs of the same circle are equal, if the related chords are congruent.

∴ arc RS ≅ arc QS ≅ are QR.

Let m(arc RS) = m(arc QS) = m(arc QR) = x      ….(1)

m(arc RS) + m(arc QS) + m(arc QR) = 360°    …..(Measure of a circle is 360°)

x + x + x = 360°       ….[From (1)]

3x = 360°

x = 360°/3

∴ x = 120°

∴ m(arc RS) = m(arc QS) = m(arc QR) = 120°

∴ m(arc QRS) = m(arc QR) + m(arc RS)     …. (Arc addition postulate)

∴ m(arc QRS) = 120° + 120°

∴ m(arc QRS) = 240°.

Chord AB ≅ chord CD  (Given)

∴ Arc ACB ≅ arc CBD       …. (Arcs corresponding to congruent chords)

∴ m(arc ACB) = m(arc CBD)        …..(1)

But m(arc ACB) = m(arc AC) + m(arc CB)        …..(2)

and m(arc CBD) = m(arc CB) + m(arc BD)         …..(3)

From (1), (2) and (3), We get,

m(arc AC) + m(arc CB) = m(arc CB) + m(arc BD)

∴ m(arc AC) = m(arc BD)

∴ arc AC ≅ arc BD.

Seg OA ≅ seg OB              ….(Radii of the same circle)

chord AB is equal to radius               ….(Given)

∴ seg AB ≅ seg OA ≅ seg OB

∴ Δ AOB is an equilateral triangle.

∴ ∠ AOB = 60°        ….. (Angle of an equilateral triangle)

The measure of an angle subtended by an arc at a point on the circle is half of the measure of the angle subtended by the arc at the centre.

∴ ∠ ACB = \(\frac{1}{2}\)∠ AOB

∴ ∠ ACB = \(\frac{60^0}{2}\)

∴ ∠ ACB = 30°

m(arc AB) = ∠ AOB               ……(Definition of measure of a minor arc)

∴ m(arc AB) = 60°

m(arc AB) + m(arc ACB) = 360°     ….. (Measure of a circle is 360°)

∴ 60° + m(arc ACB) = 360°

∴ m(arc ACB) = 360° — 60°

∴ m(arc ACB) = 300°.

Answer :

(1) ∠ AOB = 60°,

(2) ∠ ACB = 30°,

(3) m(arc AB) = 60°

(4) m(arc ACB) = 300°.

(1) □ PQRS is cyclic.

∠ PSR + ∠ PQR = 180° .(Opposite angles of cyclic quadrilateral are  supplementary)

∴  110° + ∠ PQR = 180°

∴ ∠ PQR = 180° — 110°

∴ ∠ PQR = 70°

(2) ∠ PSR = \(\frac{1}{2}\) m(arc PQR)     ……(Inscribed angle theorem)

110° =  m(arc PQR)

∴ m(arc PQR) = 110° × 2

∴ m(arc PQR) = 220°

(3) Chord PQ ≅ chord QR      …..(Given)

arc PQ ≅ arc QR     ……(Arcs corresponding to congruent chords)

Let m(arc PQ) = m(arc QR) = x

m(arc PQ) + m(arc QR) = m(arc PQR)       …..(Arcs addition postulate)

∴ x + x = 220

∴ 2x = 220

∴ x = \(\frac{220}{2}\) 

∴ x = 110°

m(arc QR) = 110° and m(arc PQ) = 110°

(4) ∠ PRQ = \(\frac{1}{2}\)  m(arc PQ)           ……(Inscribed angle theorem)

∠ PRQ = \(\frac{1}{2}\)  × 110°

∠ PRQ = 55°.

Answer :

(1) ∠ PQR = 70°,

(2) m(arc PQR) = 220°,

(3) m(arc QR) = 110°

(4) ∠ PRQ= 55°.

□ MRPN is cyclic    ….(Given)

by theorem of cyclic quadrilateral,

∠ R + ∠ N = 180°

∴ (5x — 13)° + (4x + 4)° = 180°

∴ 5x – 13 + 4x + 4 = 180

∴ 9x — 9 = 180

∴ 9x = 180 + 9

∴ 9x = 189

∴ x = \(\frac{189}{9}\) 

∴ x = 21

∠ R = 5x — 13

∠ R = 5(21) — 13

∠ R = 105 — 13,  ∴ ∠ R = 92

∠ N = (4x + 4)

∠ N = 4(21) + 4

∠ N = 84 + 4 ∴ ∠ N = 88°

Answer : ∠ R = 92° and ∠ N = 88°.

Proof : Let seg RT intersect circle at point M. Draw seg MS.

Angle inscribed in a semicircle is a right angle

∴ ∠ RMS = 90°

∴ seg SM ⊥ side RT

∴ Δ TMS is a right angled triangle.

∴ ∠ TMS + ∠ MTS + ∠ TSM = 180°      …. (Sum of all angles of a triangle is 180°)

∴ 90° + ∠ MTS + ∠ TSM = 180°

∴ ∠ MTS  + ∠ TSM = 180° — 90°

∴ ∠ MTS + ∠ TSM = 90°

∴ ∠ MTS < 90°

i.e. ∠ RTS < 90°  (R—M—T)

∴ ∠ RTS is an acute angle  (By definition)

Given : □ PQRS is rectangle.

To prove :  □ PQRS is a cyclic quadrilateral.

Proof :  □ PQRS is a rectangle.       …..(Given)

∴ ∠ P = ∠ Q = ∠ R = ∠ S = 90°      ….. (Angles of a rectangle)

∠ P + ∠ S  = 90° + 90° = 180°

∴ PQRS is a cyclic quadrilateral.    …. (By converse of cyclic quadrilateral theorem)

∠ WZP = ∠ WTP = 90°      ….(Given)

∴ ∠ WZP + ∠ WTP = 90° + 90° = 180°

∴ by converse of cyclic quadrilateral theorem,

□ WZPT is cyclic.

∠ XZY = ∠ XTY = 90°      ….(Given)

∴ points X and Y on line XY subtend equal angles at two distinct points Z and T on the same side of line XY.

If two points on a given line subtend equal angles at two distinct points which lie on the same side of the line, then the four points are concyclic.

∴ Points X, Z, T, Y are concyclic.

m(arc NS) = 125°, m(arc EF) = 37°      …(Given)

∠ NMS has its vertex in the exterior of the circle and intercepts arc EF and arc NS.

∴ ∠ NMS = \(\frac{1}{2}\) [m(arc NS) — m(arc EF)]

∴ ∠ NMS = \(\frac{1}{2}\) × [125°— 37°]

∴ ∠ NMS = \(\frac{1}{2}\) × 88°  ∴ ∠ NMS = 44°

Answer : ∠ NMS = 44°.

m(arc AE) = 95°, ∠ ABE = 108°  (Given)

∠ ABE has its vertex inside the circle and intercepts arc AE and its vertically opposite

∠ DBC intercepts arc DC.

∠ ABE = \(\frac{1}{2}\) [m(arc AE) + m(arc 130)]

∴ 108° = \(\frac{1}{2}\) × [95° + m(arc 130)]

∴ 108° × 2 = 95° + m(arc DC)

∴ 216° = 95° + m(arc DC)

∴ m(arc DC) = 216° — 95°

∴ m(arc DC) = 121°

Answer :  m(arc DC) = 121°.

Ray PQ is tangent touching the circle at point Q and line PRS is secant intersecting the circle at points R and S.

∴ PQ² = PR × PS         ….. (By tangent secant segments theorem)

∴ 12² = 8 × PS

∴ PS = \(\frac{12×12}{8}\)

∴ PS = 18

PR + RS = PS               ….(P—R—S)

∴ 8 + RS = 18

∴ RS = 18 — 8

∴ RS = 10

Answer : PS = 18 and RS = 10.

Chords MN and RS intersect each other at point D inside the circle.

∴ by theorem of internal division of chords,

DM × DN = DR × DS

∴ 8 × DN = 15 × 4

∴ DN = \(\frac{15×4}{8}\)

∴ DN = 7.5

Answer : DN = 7.5.

Let DS = x

DR + DS = RS        …..(R—D—S)

∴ DR + x = 18

∴ DR = (18 —x)

Chords MN and RS intersect each other at point D inside the circle.

∴ by theorem of internal division of chords,

DM × DN = DR × DS

∴ 9 × 8 = (18 — x) × x

∴ 72 = 18x — x²

∴ x² — 18x + 72 = 0

∴ x² — 12x— 6x + 72 = 0

x(x — 12) — 6(x — 12) = 0

(x — 12)(x — 6) = 0

Either x—12 = 0 or x – 6 = 0

∴ x = 12 or x = 6

DS = 12 or DS = 6

Answer : DS = 12 or DS = 6.

Ray AB is tangent to the circle at point B. Line ACD is secant intersecting the circle at points C and D.

∴ by tangent secant segments theorem,

AB² = AC × AD

∴ 12² = 8 × AD

∴ AD = \(\frac{12×12}{8}\)

∴ AD = 18

AC + CD = AD       ….(A—C—D)

∴ 8 + CD = 18

∴ CD = 18 — 8

∴ CD = 10

seg OE ⊥ chord CD  (Given)

Perpendicular drawn from the centre of the circle to the chord bisects the chord.

∴ DE = \(\frac{1}{2}\) × CD,  ∴ DE = \(\frac{1}{2}\) × 10 = 5

∴ DE = 5

Answer : 

(1) AD = 18,

(2) DC = 10

(3) DE = 5.

PR = 6 + 10

PR = 16

Chord ST and chord QR intersect at point P outside the circle.

∴ by theorem of external division of chords,

PS × PT = PQ × PR

∴ 8 × PT = 6 × 16

∴ PT = \(\frac{6×16}{8}\)

∴ PT = 12

PS + TS = PT      …(P—S—T)

∴ 8 + TS = 12

∴ TS = 12 — 8

∴ TS = 4

Answer : TS = 4.

Proof :

Line DF is tangent to the circle touching the circle at point F and line DGE is secant intersecting the circle at points G and E.

∴ by tangent secant segments theorem,

DF² = DG × DE      ……(1)

In Δ DFE,

∠ DFE = 90°                 …..(By tangent theorem)

∴ by Pythagoras theorem,

DE² = DF² + EF²

∴ DE² = DF² + (2r)²                 ….(∵ Diameter is twice the radius)

∴ DE² = DF² + 4r²

∴ 4r² = DE² — DF²

∴ 4r² = DE² — DG × DE                     ….[From (1)]

∴ 4r² = DE (DE — DG)

∴ 4r² = DE × GE               …(D—G—E)

∴ DE × GE = 4r².