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CIRCLE

Theorem-Radius is perpendicular

to the tangent

Sums based on theorem

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THEOREM

A tangent at any point of a circle is perpendicular to the

radius, through the point of contact.

O

A

l

seg OA ⊥ line l

Let us take point B on the tangent

B

90^o

What is ∠AOB ?

∠OAB = 90^o

OR

[Radius is perpendicular to the tangent]

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In ◻OPTQ,

∠POQ = 110º

Sol:

∠OPT

=

∠OQT

=

90º

[Radius is perpendicular

to the tangent]

∠POQ

+

∠OPT

+

360º

∠OQT

+

∠PTQ

=

[Sum of all angles of quadrilateral is 360º]

∴

110

+

90

+

360

90

+

∠PTQ

=

∴

–

360

290

∠PTQ

=

∴

70º

∠PTQ

=

T

P

Q

O

110^o

?

∴

360

290

+

∠PTQ

=

70^o

Q. If TP and TQ are the two tangents to a circle with centre O

so that ∠POQ = 110°, Find ∠PTQ.

Consider □OPTQ

We know that, sum of all angles of quadrilateral is 360º

Observe ∠OPT

∴ ∠OPT = 90º

Observe ∠OQT

∠OQT = 90º

We know that, radius is perpendicular to the tangent

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O

P

Q

In ΔOPQ ,

[Radius is perpendicular to the tangent]

OQ²

=

OP²

+

PQ²

∴

12²

=

5²

+

PQ²

∴

144

=

25

+

PQ²

∴

144

–

25

=

PQ²

[Pythagoras theorem]

∴

PQ²

=

119

∴

PQ

=

119

∴

Length of PQ is

cm.

119

Sol.

5 cm

12 cm

?

∠OPQ = 90^o

Now, let us apply Pythagoras theorem

Consider ΔOPQ

Q. 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 = 12 cm.

Find length of PQ?

Observe ∠OPQ

∴ ∠OPQ = 90º

We know that, radius is perpendicular to the tangent