Question

EXERCISE - 9.1
1. Fill in the blanks
(1) A tangent to a circle intersects it in ........ point (s).
(i) A line intersecting a circle in two points is called a
........
(ii) The number of tangents drawn at the end points of the diameter is
(iv) The common point of a tangent to a circle and the circle is called
(V) We can draw tangents to a given circle.
2. 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 so that OQ = 13 cm. Find length of PQ.
3. Draw a circle and two lines parallel to a given line such that one is a tangent and the other,
a secant to the circle.
4. Calculate the length of tangent from a point 15 cm. away from the centre of a circle of
radius 9 cm.
5. Prove that the tangents to a circle at the end points of a diameter are parallel.​

Answer 1

Answer:

1) i) 1

ii) secant line

iii) two

iv) point of contact

v) We can draw tangents to a given circle.

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2)Given,

Radius OP=5cm and OQ=12cm

PQ is the tangent to the circle.

∠OPQ=90

So,by Pythagoras theorem we get,

PQ2=OQ2-OP2

PQ2=12*12-5*5

PQ2=144-25

PQ2=119

PQ=root of 119

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3)Let the given line be AB

And circle be with centre O

Note AB II CD II EF

Here CD is a secant

(Intersecting circle at 2 points P and Q)

And

EF is a tangent

(Intersecting circle at R)

Let the given line be AB And circle be with centre O

Note AB II CD II EF

Here CD is a secant

(Intersecting circle at 2 points P and Q)

and EF is a tangent

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4)Length of tangent = x Radius =9cm

Distance to the tangent from contre = 15cm.

From the figure :-

(Distance from the centre)?-radius?+length of tangent?

152-92=length of tangent?

225-81=length of tangent 2

Length of tangent =v144-12cm.

Therefore,The length of from a point 15cm away from the centre of circle of radius 9cm is 12cm.

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5)Let AB be a diameter of the circle. Two tangents PQ and RS are drawn at points A and B respectively.

Radius drawn to these tangents will be perpendicular to the tangents.

Thus, OA ⊥ RS and OB ⊥ PQ

∠OAR = 90º

∠OAS = 90º

∠OBP = 90º

∠OBQ = 90º

It can be observed that

∠OAR = ∠OBQ (Alternate interior angles)

∠OAS = ∠OBP (Alternate interior angles)

Since alternate interior angles are equal, lines PQ and RS will be parallel.

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hope it helps you...