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.

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.

In right triangle PTO  By using Pythagoras theorem, 

PO² = OT² + TP² 

⇒ 10² = 6² + TP² 

⇒  TP²  = 100 - 36

⇒  TP²  = 64

⇒  TP  = 8

Let P be a point on AB such that, PC is at right angles to the Line Joining the centers of the circles.

Note that, PC is a common tangent to both circles.

This is because tangent is perpendicular to radius at point of contact for any circle.

let ∠PAC= α and ∠PBC = β.

PA = PC [lengths of the tangents from an external point C]

In a triangle CAP, ∠PAC = ∠ACP = α

similarly PB = CP and ∠PCB = ∠CBP = β

now in the triangle ACB,

∠CAB + ∠CBA + ∠ACB = 180°   [sum of the interior angles in a triangle]

α + β + (α + β) = 180°   (Since ∠ACB = ∠ACP + ∠PCB = α + β.

2α + 2β = 180°

α + β = 90°

∴ ∠ACB = α + β = 90°

Given- PQ is a tangent to a circle with centre O at Q. QOR is a diameter of the given circle so that ∠POR=120^o. To find out- ∠OPQ=?
Solution- QOR is a diameter.
∴OQ is a radius through the point of contact Q of the tangent PQ. ∴∠OQP=90^o since the radius through the point  of contact of a tangent to a circle is perpendicular  to the tangent.∴∠OPQ+∠OQP=120^o
(external angles of a triangle=sum of the internal opposite angles )
∴∠OPQ = 120^o − 90^o = 30^o.

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 = 

From the given figure, CD = 4 cm
CD=AD=4CD = AD = 4CD=AD=4 (tangents drawn from an exterior point are equal)
CD=BD=4CD = BD = 4CD=BD=4 (tangents drawn from an exterior point are equal)
Now, AB=AD+BD=4+4=8AB = AD + BD = 4 + 4 = 8AB=AD+BD=4+4=8 cm

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

As the lengths of tangents drawn from an external point to a circle are equal,

For the circle with center O, TQ = TP ….(i)

For the circle with center O′, TR = TP ….(ii)

Dividing equation (i) by (ii), we get,

TQ : TR = 1 : 1

The correct answer is b

∠ABC = 30 degrees

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°