Find the length of the tangent from a point (6, 1) to the circle $x^{2} + y^{2} - 4 x = 0.$

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$\sqrt{17}$

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$\sqrt{13}$

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Solution

The correct option is D
$\sqrt{13}$

Method (1)

Given circle is $x^{2} + y^{2} - 4 x = 0$

Std form $(x - 2)^{2} + y^{2} = 4$

Centre $\equiv (2, 0)$

Radius = 2 units

$∴ (T a n g e n t l e n g t h)^{2} = P C^{2} - C T^{2}$

$= {\sqrt{4^{2} + 1^{2}}}^{2} - r a d i u s^{2}$

$= 17 - 4$

$= 13$

$= \sqrt{13} u n i t s$ .

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