Question

Prove that the perpendicular at the point of contact to the tangent to a circle passes
through the centre.
The length of a tangent from a point Aat distance 5 cm from the centre of the circle is​

Answer 1

ANSWER

A chord CD is drawn which is parallel to XY and at a distance of 8cm from A.

As we know that tangent at any point of a circle is perpendicular to the radius through the point of contact.

∴∠OAY=90°

As sum of cointerior angle is 180°.

Therefore,

∠OAY+∠OED=180°

⇒∠OED=90°

AE=8cm(From fig.)

Now in △OEC, by pythagoras theorem,

OC

2

=OE

2

+EC

2

⇒EC

2

=OC

2

−OE

2

⇒EC

2

=(5)

2

−(3)

2

⇒EC=

25−9

​

=4

Therefore,

Length of chord CD=2×CE(∵perpendicular from centre to the chord bisects the chord)

⇒CD=2×4=8cm

Hence the length of the chord CD is 8cm.

Answer 2

Answer:

1. Proof of 'The perpendicular at the point of contact to the tangent to a circle passes through the center'

2. Solution of 'The length of a tangent from a point A at distance 5 cm from the center of the circle is​ 3cm'

Step-by-step explanation:

Let us assume that,

O is the center of the given circle.

A tangent PR has been drawn touching the circle at point P.

1. Draw QP ⊥ RP at point P, such that point Q lies on the circle.

∠OPR = 90°  (radius ⊥ tangent)

Also, ∠QPR = 90°  (Given)

∴ ∠OPR = ∠QPR

Now, the above case is possible only when center O lies on the line QP.

Hence, perpendicular at the point of contact to the tangent to a circle passes through the center of the circle.

AB is a tangent drawn on this circle from point A.

∴ OB ⊥ AB OA = 5cm and AB = 4 cm (Given)

In ΔABO, By Pythagoras theorem in ΔABO,

OA2 = AB2 + BO2

⇒ 52 = 42 + BO2

⇒ BO2 = 25 - 16

⇒ BO2 = 9

⇒ BO = 3

∴ The radius of the circle is 3 cm

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Thanks!