Math Olympiad Test: Circles- 4 - Free MCQ with solutions Class 10

According to question,
s = (a + b + c/2) ⇒ 2s = a + b + c
B is external point and BD and BF are tangents and from an external point the tangents drawn to a circle are equal in length
∴ BD = BF;
Similarly, AF = AE; CD = CE
s = Semi perimeter = 
⇒ 2s = AB + AC + BC
⇒ 2s = AF + FB + AE + EC + BD + DC
⇒ 2s = 2AE + 2CE + 2BD
⇒ s = AE + CE + BD
⇒ s = AC + BD ⇒ s - b = BD.

In figure, AB is a chord of circle C₁ which is a tangent to C₂.
Since, tangent is perpendicular to radius through point of contact
∴ ∠OCA = 90° ⇒ OA = a, OC = b
In ∆OCA, (OA)² = (OC)² + (AC)²
⇒ a² = b² + (AC)² ⇒ AC = 
∴ Length of chord AB = 2AC = 2 

A tangent to a circle is a line that intersects the circle at only one point. On every point on the circle, one tangent can be drawn as shown in the figure below.

As per the above diagram, we see that a circle can have infinitely many tangents.

∠BPA = 90° (Angle in semicircle)
In ∆ BPA, ∠ABP + ∠BPA + ∠PAB = 180°
⇒ 30° + 90° + ∠PAB = 180°
⇒ ∠PAB = 60°
Also, ∠POA = 2∠PBA
⇒ ∠POA = 2 × 30° = 60°
⇒ OP = AP ...(i)
(side opposite to equal angles)

In ∆OPT, ∠OPT = 90°
∠POT = 60° and ∠PTO = 30° [angle sum property of a D]
Also ∠APT + ∠ATP = ∠PAO [exterior angle property]
∴ ∠APT + 30°= 60° ⇒ ∠APT = 30°
∴ AP = AT ...(ii) (side opposite to equal angles)
From (i) and (ii), AT = OP = radius of the circle; and AB = 2r
⇒ AB = 2AT ⇒ AB/AT = 2 ⇒ AB : AT = 2 : 1

In quadrilateral YBQX
∠YXQ = 1/2 reflex ∠QBY = 1/2 x 210° = 105°
PQ is a straight line
⇒ ∠PXY + ∠YXQ = 180° (linear pair)
⇒ ∠PXY = 180° - 105° = 75°
Now, in quadrilateral PAYX
∠PXY = 1/2∠PAY
∴ ∠PAY = 2∠PXY = 150°

∠ABC = 90° [Angle in a semicircle ]
In ∆ ABC, we have
∠ACB + ∠CAB + ∠ABC = 180°

⇒ 50° + ∠CAB + 90° = 180°
⇒ ∠CAB = 40°
Now, ∠CAT = 90° ⇒ ∠CAB + ∠BAT = 90°
⇒ 40° + ∠BAT = 90°⇒ ∠BAT = 50°

(a) Draw OC perpendicular on AB,
In Δ OBC & Δ OAC
∠OCB = ∠OCA (each 90°)
OB = OA (radii of same circle)

OC is common side in both triangles
∴ ΔOBC ≅ ΔOAC (RHS congruence)

Now, in ΔOCA (OA)² = (OC)² + (CA)²

Hence, radius of circle = 5 cm
(b) In ΔOPT,
(OP)² = (OT)² + (PT)² 
⇒ (x + 4)² = x² + (8)2
⇒ x² + 16 + 8x = x² + 64
⇒ 8x = 48 ⇒ x = 6 cm
(c) PQ = 10 cm
We know, length of tangents drawn from an external point to a circle are equal.
⇒ PQ = PR ...(i)
Also, SQ = SU ...(ii)
and TU = TR ...(iii)
Now, perimeter of DPST
= PS + ST + PT = PS + SU + UT + PT
= PS + SQ + TR + PT (Using (ii) & (iii))
= PQ + PR = PQ + PQ (Using (i))
= 2 PQ = 2 × 10 = 20 cm.

Given, Chord AB = 24 cm, Radius OB = OA = 13 cm
Draw OP ⊥ AB
In D OPB, OP ⊥ AB ⇒ AP = PB
[Perpendicular from centre on chord bisect the chord] =(1/2)AB= 12
Also, OB² = OP² + PB²
⇒ (13)² = OP² + PB² ⇒ 169 = OP² + (12)²
⇒ OP² = 169 - 144 = 25 ⇒ OP = 5 cm
In Δ BPC, BC² = x² + BP² [By pythagoras theorem]
BC² = x² + 144 ...(i)
In ΔOBC, OC² = OB² + BC²
⇒ (x + 5)² = (13)² + BC² ⇒ x = 288/10=28.8cm

Put the value of x in (i), we get
BC² = x² + 144 = ((144)²/25) + 144⇒ BC = 31.2
⇒ AC = BC = 31.2 cm

Since, PQL is a tangent and OQ is a radius, so ∠OQL = 90°
∴ ∠OQS = (90°- 50°) = 40°
Now, OQ = OS ⇒ ∠OSQ = ∠OQS = 40°
Similarly, ∠ORS = (90°- 60°) = 30°
And, OR = OS ⇒ ∠OSR = ∠ORS = 30°
∴ ∠QSR = ∠OSQ + ∠OSR = (40°+ 30°) = 70°
Now, ∠ROQ = 2 ∠QSR = 140°
∠ROQ + ∠ORP + ∠OQP + ∠RPQ = 360° (Angle sum property of quadrilateral QORP)
⇒ 140° + 90° + 90° + ∠RPQ = 360°
⇒ ∠RPQ = 40°

Produce BD to meet the bigger circles at E. Join AE. Then
∠AEB = 90° [Angle in a semicircle]
OD ⊥ BE
[∵ BE is tangent to the smaller circle at D and OD is its radius] BD = DE [∵ BE is a chord of the circle and OD ^ BE]
∴ OD || AE [∵ ∠AEB = ∠ODB = 90°]
In ΔAEB, O and D are mid-points of AB and BE. Therefore, by mid-point theorem, we have

OD = 1/2AE
⇒ AE = 2 × 8 = 16 cm
In ΔODB, we have
OB² = OD² + BD² [By Pythagoras Theorem]
⇒ 169 = 8² + BD²
⇒ BD² = 169 – 64 = 105 ⇒ BD = 105 cm
⇒ DE = √105 cm [∵ BD = DE]
In ΔAED, we have
AD² = AE² + ED² [By Pythagoras Theorem]
⇒ AD² = 16² + (√105)² = 256 + 105 = 361
⇒ AD = 19 cm