The length of the tangent from an external point P to a circle of radius 5 cm is 10 cm. The distance of the point from the centre of the circle is

(a) 8 cm (b) $\sqrt{104}$ cm (c) 12 cm (d) $\sqrt{1} 25$ cm

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Solution

In ΔOAP

$∠ O A P = 90^{o}$

By Pythagoras theorem

$O P^{2} = O A^{2} + A P^{2} O P^{2} = 5^{2} + 10^{2} O P^{2} = 25 + 100 O P = \sqrt{125} c m$

Hence The distance of the point from the centre of the circle is $\sqrt{125}$ cm

correct option (d) $\sqrt{125}$

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The length of the tangent from an external point P to a circle of radius 5 cm is 10 cm. The distance of the point from the centre of the circle is (a) 8 cm (b) √104 cm (c) 12 cm (d) √125 cm

In ΔOAP ∠OAP=90o By Pythagoras theorem OP2=OA2+AP2OP2=52+102OP2=25+100OP=√125 cm Hence The distance of the point from the centre of the circle is √125 cm correct option (d) √125

The length of the tangent from an external point P to a circle of radius 5 cm is 10 cm. The distance of the point from the centre of the circle is

(a) 8 cm (b) $\sqrt{104}$ cm (c) 12 cm (d) $\sqrt{1} 25$ cm

In ΔOAP

$∠ O A P = 90^{o}$

By Pythagoras theorem

$O P^{2} = O A^{2} + A P^{2} O P^{2} = 5^{2} + 10^{2} O P^{2} = 25 + 100 O P = \sqrt{125} c m$

Hence The distance of the point from the centre of the circle is $\sqrt{125}$ cm

correct option (d) $\sqrt{125}$