Math Olympiad Test: Circles- 3 - Class 10 MCQ

Consider chord QR

∴∠QRP = ∠QSR

⇒ ∠QSR = y° [∴ QRP = y°]

⇒ ∠PSR = y°

Since angle in a semicircle is a right angle. 

∴∠QRS = 90°

Now, ∠PRS = ∠QRP + ∠QRS

⇒ ∠PRS = y° + 90°

In ΔPRS, we have

∠SPR + ∠PRS + ∠PSR = 180°

⇒ x° + y°+ 90° + y° = 180° 

⇒ x° + 2y° = 90°

Perimeter of ΔABC = AB + BC + AC
= AB + (BP + PC) + AC
= (AB + BQ) + (CR + AC)
[∵ BP = BQ, PC = CR]
= AQ + AR
(∵AQ = AR, tangents of circle from same point A)
= 2 AQ = 2 x 5 = 10 cm.

M is the mid-point of AB,

∴ AM = 6 cm

AO (r₁) = 10 cm,

AO' (r₂) = 8 cm

AB is perpendicular to OO' , then

In ΔAOM, 100 = 36 + OM² [using pythagoras theorem]

⇒ OM = 8 cm; In ΔAMO', 64 = 36 + MO'²

⇒ √28 = MO' = 2√7 = MO'

∴ OO' = (2√7 + 8) cm

x + y = 10 cm ... (i)
y + z = 12 cm    ... (ii)
x + z = 28cm    ... (iii)
Adding (i), (ii) and (iii), we get
2(x + y + z) = 50
⇒ x + y+ z = 25

∠BAT = 30° (given)
Since angle made by chord with a tangent is equal to angle made by it in alternate segment
∴ ∠APB = 30°
Now, ∠APB + ∠AQB = 180°
[Opposite angles of cyclic quadrilateral APBQ]
⇒ 30° + ∠AQB = 180° ⇒ ∠AQB = 150°
Also, ∠AOB = 60° [Angle subtended by an arc at centre is double the angle subtended by it on remaining part of circle] 
∴ ∠AOB = 60°, ∠AQB = 150°

x + y = 6 cm ... (i)

y + z = 7 cm ... (ii)

z + w = 4 cm ... (iii)

Using (ii) and (iii),

y - w = 3 cm

Now using (i) x + (3 + w) = 6 cm

⇒ x + w = 3 cm

⇒ AD = 3 cm

In the given figure, PQ is the common chord of the two circles.
⇒ AB bisects the common chord PQ at M.
∴ PM = MQ
Moreover, PQ is perpendicular to AB.
∴ Option (C) is correct.

Since OT is perpendicular to FT, Using pythagoras theorem,
100 = 64 + r²

∴ r = 6 cm

OX= OY = OZ = r    (radius of circle)
Taking ∠X = θ and ∠Z = α
Then in ΔYXO, OX = OY ⇒ ∠X = ∠OYX = θ
Now, in ΔOZY OZ= OY⇒ ∠Z= ∠OYZ = α
Now, XOZY is a quadrilateral,
⇒ ∠X+∠Y+ ∠Z+∠O = 360°
⇒ θ + θ + α + α +∠O = 360°
⇒ ∠O = 360° - 2(α + θ)
⇒ ∠O = 360° - 2 (∠OZX + ∠XZY + ∠OXZ + ∠ZXY)
⇒ ∠O = 360° - 2 (∠OZX + ∠OXZ) - 2(∠XZY + ∠ZXY)
⇒ ∠O = 360° - 2(180° - ∠O) - 2(∠XZY + ∠ZXY)
⇒ ∠O = 360° - 360° + 2∠O - 2(∠XZY + ∠ZXY)

Since PA and PB are tangents.
Also, tangent is ⊥ r to radius through point of contact,

∴ ∠PAO = 90° and ∠PBO = 90°

In quadrilateral APBO

∠APB + ∠PAO + ∠PBO + ∠AOB = 360°
80° + 90° + 90° + ∠AOB = 360°
⇒ ∠AOB = 100° ⇒ ∠AQB = 1/2∠AOB = 50°