1. If angle between two radii of a circle is 130º, then the angle between the tangents at the ends of the radii is:
(A) 90º
(B) 50º
(C) 70º
(D) 40º
Answer: (B)
Explanation: If the angle between two radii of a circle is 130º, then the angle between tangents is 180º − 130º = 50º. (By the properties of circles and tangents)
2. The length of tangent from an external point P on a circle with centre O is
(A) always greater than OP
(B) equal to OP
(C) always less than OP
(D) information insufficient
Answer: (C)
Explanation: Observe figure
The angle between tangent and radius is 90º. This implies OP is the hypotenuse for the right triangle OQP right angled at Q and hypotenuse is the longest side in aright triangle, therefore the length of tangent from an external point P on a circle with centre O is always less than OP.
3. If angle between two tangents drawn from a point P to a circle of radius ‘a’and centre ‘O’ is 90°, then OP =
(A) 2a√2
(B) a√2
(C) a/√2
(D) 5a√2
Answer: (B)
Explanation: From point P, two tangents are drawn
OT = a (given)
Also line OP bisects the RPT
Therefore,
TPO = RPO = 45º
Also
OT is perpendicular to PT
In right triangle OTP
sin 45° = OT/OP
⇒ 1/√2 = a/OP
⇒ OP = a√2
4. The length of tangent from an external point on a circle is
(A) always greater than the radius of the circle.
(B) always less than the radius of the circle.
(C) may or may not be greater than the radius of circle
(D) None of these.
Answer: (C)
Explanation: Observe the figure,
In figure OQ is the radius and QP is the tangent.
For right triangle OQP, radius and tangents are two smaller sides, smaller than hypotenuse OP.
Also the length of tangent depends upon the distance of external point from circle. Thus,
The length of tangent from an external point on a circle may or may not be greater than the radius of circle.
5. In the following figure, AT is a tangent to the circle with centre O such that OT = 4 cm and ∠OTA = 30°. Then AT is equal to
(A) 4 cm
(B) 2 cm
(C) 2√3 cm
(D) 4√3 cm
Answer: (C)
Explanation: Join OA
We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Therefore,
∠OAT = 90°
In triangle OAT
cos30° = AT/OT
√3/2 = AT/4
AT = 2√3 cm
6. At one end A of a diameter AB of a circle of radius 5 cm, tangent XAY is drawn to the circle. The length of the chord CD parallel to XY and at a distance 8 cm from A is
(A) 4 cm
(B) 5 cm
(C) 6 cm
(D) 8 cm
Answer: (D)
Explanation: Observe the figure,
Since OC = OA= radius =5cm
Therefore
OE = AE – AO
OE = 8 – 5
OE = 3 cm
In right triangle OEC
OC2 = OE2 + CE2
⇒ 52 = 32 + CE2
⇒ CE2 = 25 – 9
⇒ CE2 = 16
⇒ CE = 4
Therefore length of chord CD = 2×4 = 8cm
7. In the following figure, if O is the centre of a circle, PQ is a chord and the tangent PR at P makes an angle of 50° with PQ, then ∠POQ is equal to
(A) 100°
(B) 80°
(C) 90°
(D) 75°
Answer: (A)
Explanation: From figure using properties of circle and tangents
∠OPQ = 90° – 50°
∠OPQ = 40°
OP = OQ = radius
So (∠E) = ∠OQP = 40°
Now In ∆POQ
∠POQ = 180° – (∠OPQ + ∠OQP)
∠POQ = 180° – (40° + 40°)
∠POQ = 100°
8. In the following figure, PA and PB are tangents from a point P to a circle with centre O. Then the quadrilateral OAPB must be a
(A) Square
(B) Rhombus
(C) Cyclic quadrilateral
(D) Parallelogram
Answer: (C)
Explanation: Since tangent and radius intersect at right angle,
So,
∠OAP = ∠OBP = 90°
⇒ ∠OAP + ∠OBP = 90° + 90° = 180°
Which are opposite angles of quadrilateral OAPB
So the sum of remaining two angles is also 180°
Therefore Quad OAPB is cyclic Quadrilateral.
9. If d1, d2 (d2 > d1) be the diameters of two concentric circles and c be the length of a chord of a circle which is tangent to the other circle, then
(A) d22 = c2 + d12
(B) d22 = c2 – d12
(C) d12 = c2 + d22
(D) d12 = c2 – d22
Answer: (A)
Explanation:
Let AB be a chord of a circle which touches the other circle at C. Then ΔOCB is right triangle.
By Pythagoras theorem
10. If a chord AB subtends an angle of 60° at the centre of a circle, then angle between the tangents at A and B is:
(A) 60°
(B) 120°
(C) 80°
(D) 100°
Answer: (A)
Explanation: Observe the figure
Using properties of circles and tangents, angle between tangents is:
= 180° – 60°
= 120°
11. If two tangents inclined at an angle 60° are drawn to a circle of radius 3 cm, then length of each tangent is equal to
(A) 2√3 cm
(B) 6√3 cm
(C) 3√3 cm
(D) 3 cm
Answer: (C)
Explanation: Let P be an external point from which pair of tangents are drawn as shown in the figure given below:
Join OA and OP
Also OP is a bisector line of ∠APC
∠APO = ∠CPO = 30°
OA ⊥ AP
Therefore, in triangle OAP
tan30° = OA/AP
1/√3 = 3/AP
AP = 3√3cm
12. In the following figure, if PQR is the tangent to a circle at Q whose centre is O, AB is a chord parallel to PR and ∠BQR = 70°, then ∠AQB is equal to
(A) 20°
(B) 40°
(C) 35°
(D) 45°
Answer: (B)
Explanation: In the given figure
13. In the adjoining figure, Δ ABC is circumscribing a circle. Then, the length of BC is
(A) 7 cm
(B) 8 cm
(C) 9 cm
(D) 10 cm
Answer: (D)
Explanation: Since lengths of tangents from same external point are equal.
Therefore,
BZ=BL=4cm
Also
AZ=AM=3cm
This gives
MC= 9 – 3 = 6 cm
Similarly LC = MC = 6cm
So length of BC = BL + LC = 4cm + 6cm = 10cm
14. In the following figure, AB is a chord of the circle and AOC is its diameter such that ACB = 50°. If AT is the tangent to the circle at the point A, then BAT is equal to
(A) 65°
(B) 60°
(C) 50°
(D) 40°
Answer: (C)
Explanation: Here
∠ABC = 90 (Angle in Semicircle)
In ∆ACB
∠A + ∠B + ∠C = 180°
∠A = 180° – (90° + 50°)
∠A = 40°
Or ∠OAB = 40°
Therefore, ∠BAT = 90° – 40° = 50°
15. In the following figure, tangents PQ and PR are drawn to a circle such that ∠RPQ = 30°. A chord RS is drawn parallel to the tangent PQ, then ∠RQS.
(A) 30°
(B) 60°
(C) 90°
(D) 120°
Answer: (A)
Explanation: Since PQ = PR
(Lengths of tangents from same external point are equal)
Therefore,
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