RD Sharma Test: Circles - Class 10 MCQ

Here ∠OPQ = 90°[Angle between tangent and radius through the point of contact]
∴ OQ² = OP² + PQ² (25)² = OP² + (24)² ⇒ OP² = 625 - 576 ⇒ OP² = 49
⇒ OP = 7 cm
Therefore, the diameter = 2 x OP = 2 x 7 = 14 cm 

Since op is perpendicular to PQ, the ∠OPQ = 90°
Now, in right angled triangle OPQ,

Let AD = x and BE = y
∴ BD = 12 -  x ⇒  BE = 12 - x [BD = BE = Tangents to a circle from an external point]
⇒ y = 12 - x ⇒ x+y = 12.......(i)
Also, AF = x and CF = 10 - x and CE = 8 - y
Also, AF = x and CF = 10 — x and CE = 8 -  y
∴ 10 - x = 8 - yx - y = 2  (ii)
On solving eq. (i) and (ii), we get x = 7 and y = 5
⇒ AD = 7 cm and BE = 5cm 

Since the angle between the two tangents drawn from an external point to a circle in supplementary of the angle between the radii of the circle through the point of contact.
∴ ∠PTQ = 180^∘−110^∘ = 70^∘

∠OPQ = 90° [Angle between tangent and radius through the point of contact]

Here ∠OPQ = 90° [Tangent makes right angle with the radius at the point of contact]
∴ OQ² = OP² + PQ² (25)² = OP² + (24)²
⇒ OP2 = 676 - 576 = 49
⇒ OP = 7 cm Therefore, the radius of the circle is 7 cm 

Let ∠OAB = ∠OBA = x [Opposite angles of opposite equal radii] And ∠AOB =180° -  40° = 140°
Now, in triangle AOB,
∠OAB + ∠OBA + ∠AOB = 180°
⇒ x + x +140° = 180°
⇒ 2x = 40°
⇒ x = 20° 
∴ ∠OAB = 20°

PD + QB = PA + QA [Tangents from an external point to a circle are equal]
⇒PD + QB = PQ

Construction: Joined OP and OQ.
Here OP ⊥ AP and OC⊥AQ and PA ⊥ AQ Also AP = AQ
Therefore. APOQ is a square. ⇒ AP = OP = OQ = 5 cm

Here SQ = 6 cm, then OQ = OS = 6/2 = 3 cm [Radii] and ∠Q = 90°
Now, in triangle OQR,OR² = QR² + OQ²
⇒ OR² = (4)² + (3)² = 16 + 9 = 25
⇒ OR = 5cm 

Since OA is perpendicular to AT. then ∠OAT = 90°
⇒ ∠OAB + ∠BAT = 90°
⇒ ∠OAB + 60° = 90° ⇒ ∠OAB = 30°
∴ ∠OAB = ∠O8A = 30" [Angles opposite to radii]
∴ ∠OAB = 180° - (30° +30° ) = 120° [Angle sum property of a triangle]
∴ Reflex ∠AOB = 360° - 120° = 240°
Now, since the degree measure of an arc of a circle is twice the angle subtended by it any point of the alternate segment of the circle with respect to the arc.
∴ Reflex ∠AOB = 2∠ACB ⇒ 240° = 2∠ACB ⇒ ∠ACB = 120°

Let O be the centre. Construction: Joined OP. 

Since OP bisects ∠P, therefore, ∠APO = ∠OPB = 30° and ∠OAP = 90°

 cm Since each tangent from an external point to a circle are equal. Therefore, PA = PB = 3√3 cm 

Since Tangents from an external point to a circle are equal.
∴ PA = PB = 4 cm.
BR = CR = 5 cm CQ = AQ = 6 cm ⇒ Perimeter of ΔPQR = PQ + QR + RP
⇒ Perimeter of ΔPQR = PA + AQ + QC + CR + BR + PB  ⇒ Perimeter of ΔPQR = 4 + 6 + 6 + 5 + 5 + 4
⇒ Perimeter of Δ PQR = 30 cm 

Since Op is perpendicular to PR, then ∠OPR = 90°
∠RPQ + ∠QPO = 90°
⇒ 50° + ∠QPO = 90°
∠QPO = 40°
Now, OP = OQ (Radii of same circle]
∴ ∠OPQ = ∠OQP = 40°[Angles opposite to equal sides]
In triangle OPQ,
∠POQ + ∠OPQ + ∠OQP = 180°
∠POQ + 40° + 40° = 180°
 = ∠ POQ = 100°
 

Here ∠Q = 90° [Angle between tangent and radius through the point of contact] Now, in triangle OPQ, OP² = QO² + PQ²
⇒ OP² = (6)² + (8)² = 36 + 64 = 100
⇒ OP = 10 cm
∴ PR = OP + OR = OP + OQ [OR = OQ = Radii]
⇒ PR = 10+6 = 16cm

Since AB II PA and BQ is intersectin.
∴ ∠BQR = LQBA = 60° [Alternate angles]
And ∠BQR = ∠QAB = 60° [Alternate segment theorem] Now, in triangle AQB,
∠AQB + ∠QBA + ∠BAQ = 180°
⇒ ∠AQ,B + 60° + 60° = 180°
⇒ ∠LAQB = 60° 

Let point of contact of RS be A, point of contact of QR be B, point of contact of PQ be C and point of contact of PS be D. Also let AS = x and AR = y
Now, AS = SD = x [Tangents from an external point]

PD = 4.2 - x
But PD = PC= 4.2 - x
And QC = 6.5 - PC = 6.5 - 4.2 + x = 2.3 + x .........(i)  
Now, again, AR = BR = y [Tangents from an external point]
⇒ QB = 7.3 -  y
But QB = QC [Tangents from an external point]

Construction : Joined OA.

Since OA is perpendicular to AT. then ∠OAP = 90°
In right angled triangle OAT 

Here ∠Q = 90° and LS = 90° [Angle between tangent and radius through the point of contact]
Now, in triangle OPQ, OP² = OQ² + QP²
⇒ OP² = (3)² + (4)²
⇒ OP² = 16 + 9 = 25
⇒ OP = 5 cm Again in triangle RSO', O'R² = O'S² + FS²
⇒ O'R² = (5)² + (12)²
⇒ O'R² = 25 +144 = 169
⇒ O'R = 13 cm
∴ PR = OP+OQ+O's+OR = 5+ 3+ 5+ 13 = 26 m
 

Since OP is perpendicular to PQ, then ∠OPQ = 90°
Now, in triangle OPQ.
OQ² = OP² + PQ² ⇒ (15)² = (5)² + PQ² ⇒ PQ² = 225 -  25
⇒ PQ² = 200 ⇒ PQ = 10√2cm

Here ∠OAP = 90°
And ∠OPA = 1/2 ∠BPA [Centre lies on the bisector of the angle between the two tangents]

Now, in triangle OPA,
∠OAP + ∠OPA + ∠POA = 180°
⇒ 90° + 40° + ∠POA =180°
⇒ ∠POA = 50° 

Here SD = RD = 5 units [Tangents from an external point]
And PB = QB = 4 cm [Tangents from an external point]
∴ QC = 10 - QB = 10 - 4 = 6 units
Also QC = RC = 6 units [Tangents from an extemal point]
∴ CD = RD + RC = 5 + 6 = 11 units
Also AP = AS = 2 units [Tangents from an external point]
∴ AD = AS + DS = 2 + 5 = 7 units Therefore. Perimeter of quadrilateral ABCD = 6 + 10 + 11 + 7 = 34 units 

Here, OP = 00 = 5 cm [Radii]
And OR = PR -  OP = 8 - 5 = 3 cm Also, OA = 5 cm [Radius]
Now, in triangle AOQ, OA² = OR² + AR² ⇒ 52 = 32 + AR²
⇒AR² = 25-9= 160 AR= 4cm
Since, perpendicular from centre of a circle to a chord bisects the chord.
∴ AB = AR+ BR = 4+ 4 = 8cm