Question

A point P is 25 cm from the centre 'O' of the circle and length of the tangent drawn from 'P' to the circle is 24 cm. Find theradius of the circle.

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Answer 1

Answer:

7cm

Step-by-step explanation:

A point P is 25 cm from the centre 'O' of the circle and length of the tangent drawn from 'P' to the circle is 24 cm.

The radius, tangent and the line joining the point and centre of the circle(PO) form a right angled triangle.

In the triangle,

PO = hypotenuse = 25cm

radius = perpendicular = r

tangent = base = 24cm

using pythagoras theorem,

$h^2 = p^2 + b^2\\ \\ \Rightarrow 25^2 = r^2 + 24^2\\ \\ \Rightarrow 625 = r^2 + 576\\ \\ \Rightarrow r^2 = 625-576=49\\ \\ \Rightarrow r =\sqrt{49}\\ \\ \Rightarrow r=7\ cm$

Hence the radius of the circle 7cm

Answer 2

Answer:

Given:

• We have been given that a point P is 25 cm from the centre 'O' of the circle and length of the tangent drawn from 'P' to the circle is 24 cm.

To Find:

• We need to find the radius of circle.

Solution:

Let us draw a circle with point p such that OP = 25cm.

TP is tangent; TP = 24cm

We need to join OT such that OT is the radius of the circle.

Now, we know that tangent drawn from an external point is perpendicular to the radius at the point of contact.

Therefore, OT ⊥ PT

Now, in right △OTP we have,

(OP)^2 = (OT)^2 + (TP)^2 [By pythagoras theorem]

$= > {OT}^{2} = \sqrt{ {OP}^{2} - {TP}^{2} }$

$= \sqrt{ {25}^{2} \times {24}^{2} }$

$= \sqrt{625 - 576}$

$= \sqrt{49}$

$= 7cm$

Therefore, the radius of circle is 7cm.