Question

From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm find the radius of the circle.​

Answer 1

$\large\bold{{Question :}}$

From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm find the radius of the circle.

$\large\bold\red{\underline{Solution :-}}$

Here, O is the center of the circle.

⟼ Given :

• OQ = 25 cm
• PQ = 24 cm

⟼ To Find : We have to find the radius OP.

Since QP is tangent, OP perpendicular to QP.

(Since, Tangent is Perpendicular to Radius ⠀⠀⠀⠀⠀⠀⠀at the point of contact)

So, ∠OPQ=90°

⟼ By Applying Pythagoras Theorem :

OP² + RQ² = OQ²

OP² + (24)² = (25)²

OP² = 625 - 576

OP² = 49

OP = √49

OP = 7 cm

Hence, The Radius is 7 cm

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⠀

-MissAbhi

Answer 2

Explaination :

Here after drawing the diagram for the question we came to knew that the length of line OQ is 25 cm and length of line PQ is 24 cm.

• (Angle OP is of 90°)

So line OQ is hypotenuse , line OP is perpendicular , line PQ is base.

Let us simply apply the concept of Pythagoras theorem to find out the length of line OP.

• Base (PQ) = 24 cm
• Hypotenuse (OQ) = 25 cm
• Perpendicular (OP) = ?

$: \: \implies \: \sf{(Hypotenuse) {}^{2} \: = \: (Base) {}^{2} \: + \: (Perpendicular) {}^{2} } \\ \\ : \: \implies \: \sf{(OQ) {}^{2} \: = \: (OP) {}^{2} \: + \: (PQ) {}^{2} } \\ \\ : \: \implies \: \sf{(25) {}^{2} \: = \: (OP) {}^{2} \: + \: (24) {}^{2} } \\ \\ : \: \implies \: \sf{(OP) {}^{2} \: = \: (25) {}^{2} - \: (24) {}^{2}} \\ \\ : \: \implies \: \sf{(OP) {}^{2} \: = \: (25 \times 25) - \: (24 \times 24)} \\ \\ : \: \implies \: \sf{(OP) {}^{2} \: = \: (625) - \: (576)} \\ \\ : \: \implies \: \sf{(OP) {}^{2} \: = \: 625 - \: 576} \\ \\ : \: \implies \: \sf{(OP) {}^{2} \: = \: 49} \\ \\ : \: \implies \: \sf{OP \: = \: \sqrt{49} } \\ \\ : \: \implies \: \red{\bf{OP \: = \: 7}}$

★ Therefore,

• Radius of the circle is of 7 cm.