If the pair of tangents are drawn from origin O to the circle $x^{2} + y^{2} - 6 x - 8 y + 21 = 0$ , meets the circle at A and B , the lengths of AB is

11

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\frac{4}{5} \sqrt{21}

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\sqrt{\frac{17}{3}}

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\frac{\sqrt{21}}{5}

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Solution

The correct option is A $\frac{4}{5} \sqrt{21}$
Given equation of circle is
$x^{2} + y^{2} - 6 x - 8 y + 21 = 0$ ,

whose centre $= (3, 4)$

and radius $= \sqrt{9 + 16 - 21} = 2$

Clearly, the line $A B$ is chord of contact and its equation is

$x x_{1} + y y_{1} + g (x + x_{1}) + f (y + y_{1}) + c = 0$

Here, $(x_{1}, y_{1}) = (0, 0)$

$∴ 0 + 0 - 3 (x + 0) - 4 (y + 0) + 21 = 0$

$\Rightarrow 3 x + 4 y - 21 = 0$ ....(i)

Now, perpendicular distance from $(3, 4)$ to the line (i) is

$C M = \frac{3 (3) + 4 (4) - 21}{\sqrt{9 + 16}} = \frac{4}{5}$

$A M = \sqrt{{A C}^{2} - {C M}^{2}} = \sqrt{4 - \frac{16}{25}} = \frac{2}{5} \sqrt{21}$

$∴ A B = 2 A M = \frac{4}{5} \sqrt{21}$

Suggest Corrections

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