(a) 0
(b) 1
(c) 2
(d) Infinite
(d)Infinite
A circle has infinitely many tangents, touching the circle at infinite points on its circumference.
(a) One point
(b) Two distinct point
(c) At the circle
(d) None of the above
(a)One point
A tangent touches the circle only on its boundary and do not cross through it.
(a) One
(b) Two
(c) Three
(d) Four
(b)Two
A circle can have two parallel tangents at the most.
(a) 90º
(b) 50º
(c) 70º
(c) 40º
(c)70º
If the angle between two radii of a circle is 110º, then the angle between tangents is 180º − 110º = 70º. (By circles and tangents properties)
(a) always greater than OA
(b) equal to OA
(c) always less than OA
(d) Cannot be estimated
(c)always less than OA
Since the tangent is perpendicular to the radius of the circle, then the angle between them is 90º. Thus, OA is the hypotenuse for the right triangle OAB, which is right-angled at B. As we know, for any right triangle, the hypotenuse is the longest side. Therefore the length of the tangent from an external point is always less than the OA.
(a) 65°
(b) 60°
(c) 50°
(d) 40°
(c)50° ∠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°
(a) 60°
(b) 70°
(c) 80°
(d) 90°
(b)70°
We can see, OP is the radius of the circle to the tangent PT and OQ is the radius to the tangents TQ.
So, OP ⊥ PT and TQ ⊥ OQ
∴ ∠OPT = ∠OQT = 90°
Now, in the quadrilateral POQT, we know that the sum of the interior angles is 360°
So, ∠PTQ + ∠POQ + ∠OPT + ∠OQT = 360°
Now, by putting the respective values, we get,
⇒ ∠PTQ + 90° + 110° + 90° = 360°
⇒ ∠PTQ = 70°
(a)3cm
AB is the tangent, drawn on the circle from point A. So, OB ⊥ AB Given, OA = 5cm and AB = 4 cm Now, In △ABO, OA2 = AB2 + BO2 (Using Pythagoras theorem) ⇒ 52 = 42 + BO2 ⇒ BO2 = 25 – 16 ⇒ BO2 = 9 ⇒ BO = 3
(c)Rhombus
(a)8 From the above figure, AB is tangent to the smaller circle at point P. ∴ OP ⊥ AB By Pythagoras theorem, in triangle OPA OA2 = AP2 + OP2 ⇒ 52 = AP2 + 32 ⇒ AP2 = 25 – 9 ⇒ AP = 4 Now, as OP ⊥ AB, Since the perpendicular from the center of the circle bisects the chord, AP will be equal to PB So, AB = 2AP = 2 × 4 = 8 cm
Two concentric circles of radii a and b (a > b) are given. Find the length of the chord of the larger circle which touches the smaller circle.
a) √a^2+b^2
b) √a^2-b^2
c) 2√a^2-b^2
d) 2√a^2+b^2
c)2√a^2-b^2
a)2 cm, 3 cm, 4 cm
a)5cm
In the adjoining figure ‘O’ is the center of circle, ∠CAO = 25° and ∠CBO = 35°. What is the value of ∠AOB?
a) 120°
b) 110°
c) 55°
d) Data insufficient
a)120°
A: What is a line called, if it meets the circle at only one point? B: Collection of all points equidistant from a fixed point is \_\_\_\_\_\_. 1: Chord 2: Tangent 3: Circle 4: Curve 5: Secant Which is correct matching? a)A-2; B-4 b)A-5; B-4 c)A-4; B-1 d)A-2; B-3
d)A-2; B-3
d)10
a)cyclic quadrilateral
d)115°
In a circle of radius 5 cm, AB and AC are the two chords such that AB = AC = 6 cm. Find the length of the chord BC
a) None of these
b) 9.6cm
c) 10.8cm
d) 4.8cm
b)9.6cm
The distance between the centres of equal circles each of radius 3 cm is 10 cm. The length of a transverse tangent AB is
a) 10cm
b) 8cm
c) 6cm
d) 4cm
b)8cm
A point P is 10 cm from the center 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
a) 4cm
b) 5cm
c) None of these
d) 6cm
d)6cm
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=
a) 75°
b) 85°
c) 95°
d) 105°
d)105°
In the given figure, PAQ is the tangent. BC is the diameter of the circle. ∠BAQ = 60°, find ∠ABC :
a) 25°
b) 30°
c) 45°
d) 60°
b)30°