From a point $Q$ , the length of the tangent to a circle is $24 c m$ and the distance from the
centre is $25 c m$ . The radius of the circle is

7 c m

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12 c m

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15 c m

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24.5 c m

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Solution

The correct option is D $7 c m$
In right $Δ O T Q$
$O T = \sqrt{O Q^{2} - O T^{2}}$
$= \sqrt{25^{2} - 24^{2}} = \sqrt{625 - 576}$
$∴$ $\sqrt{49} = 7 c m$ .

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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. The radius of the circle is

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

The correct option is D 7cmIn right ΔOTQOT=√OQ2−OT2=√252−242=√625−576∴ √49=7cm.

From a point $Q$ , the length of the tangent to a circle is $24 c m$ and the distance from the
centre is $25 c m$ . The radius of the circle is

The correct option is D $7 c m$
In right $Δ O T Q$
$O T = \sqrt{O Q^{2} - O T^{2}}$
$= \sqrt{25^{2} - 24^{2}} = \sqrt{625 - 576}$
$∴$ $\sqrt{49} = 7 c m$ .