Home / Class 10 Math / Ch 10 Circles- MCQ Online Test 1 Class 10 Maths
Chapter 10 Circles- MCQ Online Test 1 Class 10 Maths
A point P is 10 cm from the centre of a circle. The length of the tangent drawn from P to the circle is 8 cm. The radius of the circle is equal to 11 cm 10 cm 6 cm 8 cm
2. A point P is 25 cm from the centre of a circle. The radius of the circle is 7 cm and length of the tangent drawn from P to the circle is x cm. The value of x = 14 cm 20 cm 24 cm 22 cm
3. In fig, O is the centre of the circle, CA is tangent at A and CB is tangent at B drawn to the circle. If ∠ACB = 75°, then ∠AOB =
125º 115º 105º 100º
4. In figure 10.75, PA and PB are the two tangents drawn to the circle. 0 is the centre of the circle. A and B are the points of contact of the tangents PA and PB with the circle. If ∠OPA = 35°, then∠ POB =
55° 105° 95° 90°
5. In fig, O is the centre of the circle. PQ is tangent to the circle and secant PAB passes through the centre O. If PQ = 5 cm and PA = 1 cm, then the r adius of the circle is.
21 cm 22 cm 12 cm 15 cm
6. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q such that OQ = 12 cm. Length PQ is 10 cm 8.5 cm √119 cm 15 cm
7. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is 2 cm 11 cm 9 cm 7 cm
8. The length of the tangent from a point A at a circle, of radius 3 cm, is 4 cm. The distance of A from the centre of the circle is 12 cm 9 cm 5 cm 15 cm
9. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80° then ∠POA is equal to 40° 50° 70° 85°
10. If TP and TQ are two tangents to a circle with centre O so that ∠POQ = 110°, then, ∠PTQ is equal to 70° 60° 90° 50°
Class 10 Circles Quiz -1
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Question 1 of 10
1. Question
A point P is 10 cm from the centre of a circle. The length of the tangent drawn from P to the circle is 8 cm. The radius of the circle is equal to
Correct
We know that, A line drawn through the centre to the point of contact is perpendicular on it.
∠OTP = 90°
Now, in right ΔOPT
OP2 = OT2 + PT2 (10)2 = OT2 + (8)2 ⇒ 100 = OT2 + 64
⇒ OT2 = 100 – 64
⇒ OT2 = 30
⇒ OT = 6 cm
Hence, Radius of the circle is 6 cm.
Incorrect
We know that, A line drawn through the centre to the point of contact is perpendicular on it.
∠OTP = 90°
Now, in right ΔOPT
OP2 = OT2 + PT2 (10)2 = OT2 + (8)2 ⇒ 100 = OT2 + 64
⇒ OT2 = 100 – 64
⇒ OT2 = 30
⇒ OT = 6 cm
Hence, Radius of the circle is 6 cm.
Question 2 of 10
2. Question
A point P is 25 cm from the centre of a circle. The radius of the circle is 7 cm and length of the tangent drawn from P to the circle is x cm. The value of x =
Correct
Given- O is the centre of a circle to which a tangent PT=x has been drawn to the circle at T when OP=25cm. The radius of the given circle=7cm
To find: x=?
We join OT. ∴OT is a radius of the circle through the point of contact T of the tangent PT. We know that the radius through the point of contact of a tangent to a circle is perpendicular to the tangent. ∴OT⊥PT⟹∠OTP=90 °.
∴ΔOTP is a right one with OP as hypotenuse. So, applying Pythagoras theorem, we get
The tangent to the given circle PT=24cm.
Incorrect
Given- O is the centre of a circle to which a tangent PT=x has been drawn to the circle at T when OP=25cm. The radius of the given circle=7cm
To find: x=?
We join OT. ∴OT is a radius of the circle through the point of contact T of the tangent PT. We know that the radius through the point of contact of a tangent to a circle is perpendicular to the tangent. ∴OT⊥PT⟹∠OTP=90 °.
∴ΔOTP is a right one with OP as hypotenuse. So, applying Pythagoras theorem, we get
The tangent to the given circle PT=24cm.
Question 3 of 10
3. Question
In fig, O is the centre of the circle, CA is tangent at A and CB is tangent at B drawn to the circle. If ∠ACB = 75°, then ∠AOB =
Correct
OA&OB are radii drawn from O to A&B respectively.
∴∠OBC = 90° = ∠OAC
Since the radius through the point of contact of a tangent to a circle is perpendicular to the tangent.
Now, considering the quadrilateral AOBC, we have
∠OBC+∠OAC+∠ACB+∠AOB = 360° (by angle sum property of quadrilateral)
⟹90°+90°+75°+∠AOB = 360°
⟹∠AOB = 105°.
Incorrect
OA&OB are radii drawn from O to A&B respectively.
∴∠OBC = 90° = ∠OAC
Since the radius through the point of contact of a tangent to a circle is perpendicular to the tangent.
Now, considering the quadrilateral AOBC, we have
∠OBC+∠OAC+∠ACB+∠AOB = 360° (by angle sum property of quadrilateral)
⟹90°+90°+75°+∠AOB = 360°
⟹∠AOB = 105°.
Question 4 of 10
4. Question
In figure 10.75, PA and PB are the two tangents drawn to the circle. 0 is the centre of the circle. A and B are the points of contact of the tangents PA and PB with the circle. If ∠OPA = 35°, then∠ POB =
Correct
Since the radius through the point of contact of a tangent to a circle is perpendicular to the tangent.
Also, PA=PB since the lengths of the tangents, drawn from a point to a circle, are equal.
So, between ΔPOB & ΔPOA, we have
PA=PB,
PO common,
∠OAP=∠OBP.
therefore, ΔPOB≅ΔPOA
⟹∠BPO=∠APO=35°.(by SAS test) .
So, in ΔPOB, we have ∠POB = 180°−90°−35°=55°.
Incorrect
Since the radius through the point of contact of a tangent to a circle is perpendicular to the tangent.
Also, PA=PB since the lengths of the tangents, drawn from a point to a circle, are equal.
So, between ΔPOB & ΔPOA, we have
PA=PB,
PO common,
∠OAP=∠OBP.
therefore, ΔPOB≅ΔPOA
⟹∠BPO=∠APO=35°.(by SAS test) .
So, in ΔPOB, we have ∠POB = 180°−90°−35°=55°.
Question 5 of 10
5. Question
In fig, O is the centre of the circle. PQ is tangent to the circle and secant PAB passes through the centre O. If PQ = 5 cm and PA = 1 cm, then the radius of the circle is
Correct
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 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.
Incorrect
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 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.
Question 6 of 10
6. Question
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q such that OQ = 12 cm. Length PQ is
Correct
By Pythagoras theorem,
122 =52 +x2
144 =25 + x2
144 – 25 = x2
119 = x2
√119 = x
Incorrect
By Pythagoras theorem,
122 =52 +x2
144 =25 + x2
144 – 25 = x2
119 = x2
√119 = x
Question 7 of 10
7. Question
From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is
Correct
Given that,
OQ = 25cm and PQ = 24 cm
As the radius is perpendicular to the tangent at the point of contact,
Therefore, OP ⊥ PQ
Applying Pythagoras theorem in ΔOPQ, we obtain
OP2 + PQ2 = OQ2
OP2 + 242 = 252
OP2 = 625 − 576
OP2 = 49
OP = 7
Therefore, the radius of the circle is 7 cm
Incorrect
Given that,
OQ = 25cm and PQ = 24 cm
As the radius is perpendicular to the tangent at the point of contact,
Therefore, OP ⊥ PQ
Applying Pythagoras theorem in ΔOPQ, we obtain
OP2 + PQ2 = OQ2
OP2 + 242 = 252
OP2 = 625 − 576
OP2 = 49
OP = 7
Therefore, the radius of the circle is 7 cm
Question 8 of 10
8. Question
The length of the tangent from a point A at a circle, of radius 3 cm, is 4 cm. The distance of A from the centre of the circle is
Correct
We know that the tangent drawn from an external point to the circle makes right with the radius at the point of intersection with circle.
Let O= center of circle
A= External Point from which tangent is drawn
R = Point of intersection of radius and tangent.
Then, ΔARO is a right triangle.
Given, we have
OR = 3 cm
AR= 4 cm
To find: AO
By Pythagoras theorem, we have
AO²= AR²+OR²
AO²= 4²+ 3² = 16+9=25
AO²= 25
⇒ AO=5 cm [Take square root on both sides]
Incorrect
We know that the tangent drawn from an external point to the circle makes right with the radius at the point of intersection with circle.
Let O= center of circle
A= External Point from which tangent is drawn
R = Point of intersection of radius and tangent.
Then, ΔARO is a right triangle.
Given, we have
OR = 3 cm
AR= 4 cm
To find: AO
By Pythagoras theorem, we have
AO²= AR²+OR²
AO²= 4²+ 3² = 16+9=25
AO²= 25
⇒ AO=5 cm [Take square root on both sides]
Question 9 of 10
9. Question
If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80° then ∠POA is equal to
Correct
PA & PB are tangent and O is the centre.
⇒ ∠OAP = ∠OBP = 90°
PA and PB are inclined to each other at 80°
⇒ ∠APB = 80°
in Quadrilateral OAPB
∠AOB + ∠OAP + ∠APB + ∠OBP = 360°
⇒ ∠APB + 90° + 80° + 90° = 360°
⇒ ∠APB = 100°
∠POA = ∠POB = (1/2) ∠APB
⇒ ∠POA = (1/2) 100°
⇒ ∠POA = 50°
Incorrect
PA & PB are tangent and O is the centre.
⇒ ∠OAP = ∠OBP = 90°
PA and PB are inclined to each other at 80°
⇒ ∠APB = 80°
in Quadrilateral OAPB
∠AOB + ∠OAP + ∠APB + ∠OBP = 360°
⇒ ∠APB + 90° + 80° + 90° = 360°
⇒ ∠APB = 100°
∠POA = ∠POB = (1/2) ∠APB
⇒ ∠POA = (1/2) 100°
⇒ ∠POA = 50°
Question 10 of 10
10. Question
If TP and TQ are two tangents to a circle with centre O so that ∠POQ = 110°, then, ∠PTQ is equal to
Correct
∠P = 90° (Radius of the circle perpendicular to the tangent)
∠Q = 90° (Radius of the circle perpendicular to the tangent)
Given ∠O = 110°
∠T+∠P+∠Q+∠O =360° (Sum of all angles in quadrilateral PTQO)
∠T+290° =360°
∠T =360°-290°
∠T+90°+90°+110° =360°
∠T=70°
Therefore ∠PTQ= 70°
Incorrect
∠P = 90° (Radius of the circle perpendicular to the tangent)
∠Q = 90° (Radius of the circle perpendicular to the tangent)
Given ∠O = 110°
∠T+∠P+∠Q+∠O =360° (Sum of all angles in quadrilateral PTQO)
