Practice Test: Circles - Class 10 MCQ

ΔPTO is a right angled triangle at T ,where the tangent touches the circle.
So applying pythagoras theorem,
H²=P²+B²
10²=B²+8²
B= 
Radius is 6cm.

The distance from the center (O) to the external point (P) is OP = 25 cm.

The radius of the circle (OT) is 7 cm. 2.

Apply the Pythagorean theorem: OP² = OT² + PT².

Therefore, the length of the tangent from point P to the circle is 24 cm.

Explanation:

∠OAC = ∠OBC = 90°

∠OAC + ∠OBC + ∠ACB + ∠AOB = 360° ... (sum of angles of a quadrilateral)

90° + 90° + 75° + ∠AOB = 360°

∠AOB = 105°

AOB is the diameter of the circle since POB passes through O.
Now PA is the segment of the secant POB outside the circle.
So, by the tangent–secant rule, the square of the tangent = secant segment × segment of the secant outside the circle when a secant and tangent to a circle are drawn from a point outside the circle.
PB × PA = PQ × PA ⇒ PB = (PQ²) / PA
PB = (25²) / 1 = 625 / 1 = 25 cm
The diameter AOB = PB - PA = (25 - 1) cm = 24 cm
The radius = x = AOB / 2 = 24 cm / 2 = 12 cm

In the given problem, PQ is a tangent to the circle at point P, and OP is the radius to the point of tangency. By the property of tangents, OP is perpendicular to PQ, i.e., OP ⟂ PQ.

Given:

• Radius, OP = 5 cm
• OQ = 12 cm

Since OP ⟂ PQ, triangle OPQ is a right-angled triangle with the right angle at P.

Applying the Pythagorean theorem in triangle OPQ:

OQ² = OP² + PQ²

Substituting the known values:

12² = 5² + PQ²

144 = 25 + PQ²

PQ² = 144 - 25 = 119

Therefore, PQ = √119 cm.

Therefore, the radius drawn to these tangents will be perpendicular to the tangents.

Thus, OA ⊥ PA and OB ⊥ PB

∠OBP = 90°

∠OAP = 90°

In △OBP,

Sum of all interior angles = 360°

∠OAP + ∠APB + ∠PBO + ∠BOA = 360°

⇒ 90° + 80° + 90° + ∠BOA = 360°

⇒ 260° + ∠BOA = 360°

⇒ ∠BOA = 360° - 260°

⇒ ∠BOA = 100°

In △OPB and △OPA,

AP = BP (Tangents from a point)

OA = OB (Radii of the circle)

OP = OP (Common side)

Therefore, △OPB ≅ △OPA (SSS congruence criterion)

And thus, ∠POB = ∠POA

The tangent at any point of a circle is perpendicular to the radius at the point of contact

In the above figure, OPTQ is a quadrilateral and ∠P and ∠Q are 90°

The sum of the interior angles of a quadrilateral is 360°.

Therefore,  in OPTQ,

∠Q + ∠P + ∠POQ + ∠PTQ = 360°

90° + 90° + 110° + ∠PTQ = 360°

290° + ∠PTQ = 360°

∠PTQ = 360° - 290°

∠PTQ = 70°

In right triangle PTO  By using Pythagoras theorem, 

PO² = OT² + TP² 

⇒ 10² = 6² + TP² 

⇒  TP²  = 100 - 36

⇒  TP²  = 64

⇒  TP  = 8

∠POQ = 180° - 120° = 60°

In ΔOPQ, we know

∠POQ + ∠OQP + ∠OPQ = 180° ...(Sum of the angles of a Δ is 180°)

60° + 90° + ∠OPQ = 180°

∠OPQ = 180° - 150° = 30°

∠OPQ = 30°

Let P be the external point and PA and PB be the tangents and OA and OB be the radii.
So OP is the hypotenuse=8cm 
Applying Pythagoras theorem,
H² = P² + B²
64 = AP² + 36
AP = 

Let ABCD be a cyclic quadrilateral having diagonals BD and AC, intersecting each other at point O.

(Consider BD as a chord)
∠BCD + ∠BAD = 180 (Cyclic quadrilateral)

∠BCD = 180− 90 = 90
(Considering AC as a chord)

∠ADC + ∠ABC = 180 (Cyclic quadrilateral)

90+ ∠ABC = 180
∠ABC = 90

Each interior angle of a cyclic quadrilateral is of 90.Hence it is a rectangle.

To prove: PQRS is a cyclic quadrilateral. 

Proof: In △ARB, we have

1/2∠A + 1/2∠B + ∠R = 180°   ....(i)   (Since, AR, BR are bisectors of ∠A and ∠B)

In △DPC, we have 

1/2∠D + 1/2∠C +  ∠P = 180°  ....(ii)   (Since, DP,CP are bisectors of ∠D and ∠C respectively)

Adding (i) and (ii),we get

1/2∠A + 1/2∠B + ∠R + 1/2∠D + 1/2∠C + ∠P = 180° + 180°

∠P + ∠R = 360° - 1/2(∠A  + ∠B + ∠C  + ∠D)

∠P + ∠R = 360° - 1/2 x 360° = 360° - 180°

⇒ ∠P + ∠R = 180°

As the sum of a pair of opposite angles of quadrilateral PQRS is 180°. Therefore, quadrilateral PQRS is cyclic.

Given, PQ = 15 cm

PR = 25 cm

We know that, PQ² = PN × PR
15² = PN × 25
225 = PN × 25
PN = 9

Answer: (c) 50°

 

∠ABC = 90° (angle in Semicircle is right angle)
In ∆ACB
∠A + ∠B + ∠C = 180°
∠A = 180° – (90° + 50°)
∠A = 40°
Or ∠OAB = 40°
Therefore, ∠BAT = 90° – 40° = 50°