The common point of the tangent and the circle is called

1

(a)

point of contact.

The length of the tangents drawn from an external point to a circle are _____________

1

(a)

equal

If a line and a circle have no point common, then the line lies ______________

1

(a)

outside the circle.

At the point of contact, angle between the tangent and the radius is is ___________

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(a)

A right angle \(\left( { 90 }^{ \circ } \right) \).

A line segment drawn through the end of a radius and perpendicular to it, is a ___________to the circle.

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(a)

tangent line.

The length of tangents drawn from an external point to a circle are equal.

1

(a) False

(b) True

The distance between two parallel tangents drawn to a circle is equal to the diameter of the circle.

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(a) False

(b) True

A tangent line intersects the circle in two points.

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(a) True

(b) False

The parallelogram circumscribing a circle is a rectangle.

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(a) True

(b) False

The length of the tangent is the length of the segment from an external point to the point of contact.

1

(a) False

(b) True

Find the length of the tangent drawn from a point whose distance from the centre of a circle is 35 cm.Given that radius of the circle is 7cm.

1

(a)

Let O is the centre of the circle and P is a point such that OP = 25 cm and PQ is the tangent to the circle. 
OQ = radius = 7 cm
In \(\triangle\)OQP, we have \(\angle \)Q = 90°
OP² = OQ² + PQ²
⇒ (25)² = 7² + PQ² ⇒ PQ² = 625 - 49 = 576
⇒ PQ = 24 cm

What is the angle between a tangent to a circle and the radius through the point of contact? Justify your answer

1

(a)

90⁰.Because radius through point of contact of tangent to a circle is perpendicular to the tangent.

What is the distance between two parallel tangents of a circle of radius 7cm?

1

(a)

parallel tangents of a circle can be drawn only at the end points of the diameter 
⇒  I₁ || I₂
⇒  Distance between I₁ and I₂ = AB = Diameter of the circle 
= 2 x r = 2 x 7 cm = 14 cm 

In the figure given below, find \(\angle QSR.\)

1

(a)

In the figure given below, find \(\angle \)QSR. 
Given: PQ and PR are tangents to a circle with centre O and 
\(\angle \)QPR = 50
To find: QSR 
sol. \(\angle \)QOR + \(\angle \)QPR = 180°
⇒ \(\angle \)QOR + 50° = 180°
⇒ \(\angle \)QOR = 130QOR
⇒ \(\angle \)QOR = \(1\over2\) \(\angle \)QSR    [Degree measure theorem]
⇒  \(\angle \)QSR =  \(1\over2\) x 130° = 65°

How many tangents parallel to a secant can a circle have?

1

(a)

2 tangents