PE 1 - Circles

CP = PD = 30/2 = 15 cm

EP = PF = 24/2 = 12 cm

∴ CE = FD = 15 – 12 = 3 cm

In ∆OPC, R² = OP² + PC²   …(1)

In ∆OPF, r² = OP² + PF²   …(2)

(1) - (2)

R² – r² = 15² - 12² = 225 – 144 = 81   …(3)

Now, in ∆OAQ

AO² = AQ² + OQ² 

⇒ AQ² = AO² – OQ²

AQ² = R² – r² = 81

⇒ AQ = 9

⇒ AB = 2 × 9 = 18

Hence, option (a).

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Let the radius of bigger circle (center X) be R cm.

Now, ∠QCY = ½ ∠BCA = 30°

∴ ∆CYQ is a 30°-60°-90° triangle

YQ = 2 ⇒ CY = 4 and CQ = 2√3

∆CXP is also a 30°-60°-90° triangle

∴ CX = 2 XP

⇒ (R + 2 + 4) = 2R

⇒ R = 6 cm

Hence, option (b).

OB = 6

∆ ABO is a 30°-60°-90° triangle

∴ AB = OB√3 = 6√3 cm

∴ $\sqrt{A{B}^2+A{C}^2}$ = $\sqrt{{\left(6\sqrt{3}\right)}^2+{\left(6\sqrt{3}\right)}^2}$ = 6√6 cm

Hence, option (a).

In ΔPRQ

∠Q + ∠R + ∠P = 180°

⇒ 180 – 2x + 180 – 2y + 45 = 180

⇒ x + y = 225/2 = 112.5

In ΔQOR

∠OQR + x + y = 180°

⇒ ∠OQR = 180 – (x + y) = 180 – 112.5 = 67.5°

Hence, option (a).

PQ = 4 + 4 + 4 + 4 = 16 cm

PA = 4 + 4 + 4 = 12 cm

In ∆ASR, PA² = AS² + PS²

⇒ 144 = 16 + PS²

⇒ PS = 8√2

Now, ∆PQR ~ ∆PSA

⇒ $\frac{RQ}{AS}=\frac{QP}{PS}$

⇒ $\frac{RQ}{4}=\frac{16}{8\sqrt{2}}$

⇒ RQ = 4√2 cm

Hence, option (c).

Direct Common Tangent = $\sqrt{d^2-{(R-r)}^2}$ = $\sqrt{{10}^2-{(4-3)}^2}$ = √99

Transverse Common Tangent = $\sqrt{d^2-{(R+r)}^2}$ = $\sqrt{{10}^2-{(4+3)}^2}$ = √51

∴ Required ratio = √99 : √51 = √33 : √17

Hence, option (b).

In ∆AOB, AB = 5 = OA = PB

∴ ∆AOB is an equilateral triangle.

⇒ ∠AOB = 60°

In quadrilateral AOBP,

∠A = ∠B = 90°

∴ ∠AOB + ∠A + ∠B + ∠P = 360°

⇒ 60 + 90 + 90 + ∠P = 360

⇒ ∠P = 120°

Since PA || OC and PB || OD

⇒ ∠COD = ∠APB = 120°

Hence, option (b).

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Considering minor arc AB, ∠APB = ½ ∠AOB = 32°

Also, ∠PAQ = 90° (Since PQ is the diameter)

In ∆PAX,

∠PAQ + ∠AXP + ∠XPA = 180°

⇒ 90 + ∠AXP + 32 = 180

⇒ ∠AXP = 58°.

Hence, option (b).

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∆AEB ≅ MED

∴ MD = AB

∆BFA ≅ NFC

∴ NC = AB

⇒ DM = AB = NC

⇒ DC = DM + MN + NC = 3AB

Now, ∆AOB ≈ ∆COD

∴ $\frac{Area(∆AOB)}{Area(∆COD)}$ = ${\left(\frac{1}{3}\right)}^2$ = $\frac{1}{9}$

Hence, option (b).

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Let C₁, C₂, & C₃ be three centers of three circles of radius a, b & c respectively PQ = 2$\sqrt{ac}$ (Length of direct common tangent when two circles touch each other externally)

Similarly,

QR = 2$\sqrt{bc}$, PR = 2$\sqrt{ab}$

PR = QR + PQ

2$\sqrt{ab}$ = 2$\sqrt{bc}$ + 2$\sqrt{ac}$

Divide by √abc on both sides.

$\frac{1}{\sqrt{c}}=\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{a}}$

Hence, option (d).