ML Aggarwal Solutions Class 10 Maths Chapter 15 Circles

In a circle with center O and radius 3cm and p is at a distance of 5cm.

That is OT = 3 cm,OP = 5 cm

OT is the radius of the circle

OT ⊥ PT 

Now in right ∆ OTP, by Pythagoras axiom,

OP²  = OT²  + PT²

(5)²  = (3)2  + PT² 

 PT²   = (5)2  – (3)2  = 25 – 9 = 16 = (4)² 

PT = 4 cm. 

Hence, the given values PT = 4 cm are not roots of the equation.

 

CT is the radius 

CP = 13 cm and tangent PT = 12 cm

 

CT is the radius and TP is the tangent

CT is perpendicular TP

Now in right angled triangle CPT,

CP²  = CT²  + PT² [using Pythagoras axiom]

OB²  = OP²  + PB² 

13²  = 5²  + PB² 

169 = 25 + PB² 

PB²   = 169 – 25

= 144

PB = 12 cm

But P is the mid-point of AB.

AB = 2PB

= 24 cm

 

Radii of the circles are 5 cm and 2.8 cm.

i.e. OP = 5 cm and CP = 2.8 cm.

(i) When the circles touch externally, then the distance between their centers = OC = 5 + 2.8

= 7.8 cm. 

(ii) When the circles touch internally, then the distance between their centers = OC = 5.0 – 2.8

= 2.2 cm 

 

(a) From A,AP and AQ are the tangents to the circle

∴ AQ = AP = 4cm 

But AC =12 cm

CQ = 12 – 4 = 8 cm.

From B,BP and BR are the tangents to the circle

BR = BP = 6 cm.

Similarly, from C,

CQ and CR the tangents

CR = CQ = 8 cm 

x = BC = BR + CR = 6 cm + 8 cm = 14 cm 

(b) From C,CR and CS are the tangents to the circle.

 

CS = CR = 3 cm.

But BC = 7 cm.

BS = BC – CS = 7 – 3 =4 cm.

Now from B,BP and BS are the tangents to the circle.

BP = BS = 4 cm 

From A,AP and AQ are the tangents to the circle.

AP = AQ = 5cm

x = AB = AP + BP = 5 + 4

= 9 cm

 

 

 

CE is tangent at D

AP = 15 cm 

PA and PB are tangents to the circle

AP = BP = 15 cm 

Similarly EA and ED are tangents

EA = ED 

Similarly BC = CD

Now perimeter of triangle PEC,

= PE + EC + PC

= PE + ED + CD + PC

PE + EA + CB + PC

(ED = EA and CB = CD)

=AP + PB = 15 + 15

= 30 cm.

(a) Let the circle touch the sides BC,CA and AB of the right triangle ABC at points D,E and F respectively,

 

where BC = a,CA = b

 

and AB = c (as showing in the given figure).

As the lengths of tangents drawn from an

External point to a circle are equal

AE = AF,BD = BF and CD = DE

OD ⊥ BC and OE ⊥ CA 

(tangents is ⊥ to radius)

OD ⊥ BC and OE ⊥ CA

(tangents is ⊥ to radius)

ODCE is a square of side r

DC = CE = r 

AF = AE = AC – EC = b – r and

BF = BD = BC – DC = a – r 

Now, AB = AF + BF

C = (b – r) + (a – r) 

2r = a + b – c 

r = a + b – c/2 

OP²  = 400 + 169

OP = a-√569 cm

Hence, the given values OP = a-√569 cm are not roots of the equation.

 

We have given that:

Three circles with centers A,B and C touch each other externally at P,Q and R respectively and the radii of these circles are

2 cm,3 cm and 4 cm.

We have given that:

(a) triangle ABC is an equilateral triangle

Each angle = 60°

∠A = 60°

But ∠A = ∠D

(Angles in the same segment)

∠D = 60° 

Now ABEC is a cyclic quadrilateral,

∠A = ∠E = 180°

60°  + ∠E = 180°

60°  + ∠E = 180°  (∠E = 180°  – 60°

∠E = 120°

Hence ∠BDC = 60° and ∠BEC = 120°

1.AB is diameter of circle with centre O.

OD ⊥ AB and C is a point on arc DB.

 

In ∆AOD,∠AOD = 90°

OA = OD (raddi of the semi – circle)

∠OAD = ∠ODA 

But ∠OAD + ∠ODA = 90°

∠OAD + ∠ODA = 90°

2∠OAD = 90°

∠OAD = 90°/2 = 45°

Or ∠BAD = 45°

(ii) Arc AD subtends ∠AOD at the centre and

∠ACD at the remaining part of the circle

∠AOD = 2 ∠ACD

90°  = 2 ∠ACD (OD ⊥ AB)

∠ACD = 90°/2 = 45°

∠ADC + ∠ABC = 180°

118°  + ∠ABC = 180°

∠ABC = 180°  – 118°  = 62°

But in ∆AEB

∠AEB = 90° 

(Angles in a semi-circle)

∠EAB = ∠ABE (AE = BE) 

∠EAB + ∠ABE = 90°

∠EAB = 90°  × ½ = 45°

∠CBE = ∠ABC + ∠ABE

= 62°  + 45°  = 107°

But AEBD is a cyclic quadrilateral

∠CAE + ∠CBE = 180°

∠CAE + 107°  = 180°

∠CAE = 180°  – 107°  = 73°

(b) AB is the diameter of semi-circle ABCDE

With center O.AE = ED and ∠BCD = 140°

In cyclic quadrilateral EBCD.

We have given that:

(a) Given: O is the center of the circle. To Prove : ∠AOC = 2 (∠ACB + ∠BAC). Proof: In ∆ABC, ∠ACB + ∠BAC + ∠ABC = 180° (Angles of a triangle) 

∠ABC = 180°  – (∠ACB + ∠BAC)….(i)

In the circle, arc AC subtends ∠AOC at

The center and ∠ABC at the remaining part of the circle.

Reflex ∠AOC = 2 ∠ABC …(ii)

Reflex AOC = 2 { (180°  – (ACB + BAC)}

But ∠AOC = 360°  – 2(∠ACB + ∠BAC)

But ∠AOC = 360°   – reflex ∠AOC

=360 – 360°  – 2(∠ACB + ∠BAC)

= 360°  – 360°  + 2 (∠ACB + ∠BAC)

=2 (∠ACB + ∠BAC)

Hence ∠AOC = 2 (∠ACB + ∠BAC)

(b) Given : in the figure, O is the center of the circle

To Prove : × + y = z.