Question

Q23. The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4
cm. Find the radius of the circle.​

Answer 1

☆ Solution ☆

Given :-

• The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm.

To Find :-

• The radius of the circle.

Step-by-Step-Explaination :-

As we know that :-

Tangent is perpendicular to the radius .

Therefore,

ABO = 90° and ABO is a right angled triangle.

Now,

Using pythogoras theorem in ABO

We have :-

AO² = BO² + AB²

=> BO² = AO² - AB²

=> BO² = 25 - 16

=> BO² = 9

=> BO = √9

=> BO = 3 cm

Hence,

Radius of the circle is 3 cm .

Answer 2

Given:

The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4cm.

So, a triangle ABO is forming from the centre of the circle.

To find :

The radius of the circle.

Solution:

Radius of the circle = OB

As we know that,

A tangent of circle is perpendicular to the radius of the circle at that same point.

So, $\angle$ABO = 90° (As given)

In ΔABO,

OA = hypotenuse = 5cm

AB = base = 4cm

OB = perpendicular = ?

According to pythagoras theoram,

$\rm \bf (H)^2 = (B)^2 + (P)^2$

Where,

H = Hypotenuse of the Δ

P = Perpendicular side of the Δ

B = Base of the Δ

So, by applying pythagoras theoram in Δ ABO,

$\rm \therefore (OA)^2 = (AB)^2 + (OB)^2$ $\textrm{So, On Putting the values}$ $\implies \rm (5cm)^2 = (4cm)^2 + (OB)^2 \\ \\ \implies \rm 25cm^2 = 16cm^2 + (OB)^2 \\ \\ \implies \rm 25cm^2 - 16cm^2 = (OB)^2 \\ \\ \implies \rm 9cm^2 = (OB)^2 \\ \\ \implies \rm OB = \sqrt{9cm^2} \\ \\ \implies \rm \bf \red{OB = 3cm}$

Hence, Length of the radius of the circle is $\sf \bf \red{3cm}$.