If OA and OB are the tangents to the circle x | vedclass

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(B) The equation of the circle is $x^2 + y^2 - 6x - 8y + 21 = 0$. The center is $C(3, 4)$ and the radius $r = \sqrt{3^2 + 4^2 - 21} = \sqrt{9 + 16 - 2

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

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(B) The equation of the circle is $x^2 + y^2 - 6x - 8y + 21 = 0$. The center is $C(3, 4)$ and the radius $r = \sqrt{3^2 + 4^2 - 21} = \sqrt{9 + 16 - 21} = \sqrt{4} = 2$.The equation of the chord of contact $AB$ from the origin $O(0, 0)$ is given by $T = 0$:$x(0) + y(0) - 3(x + 0) - 4(y + 0) + 21 = 0$$3x + 4y - 21 = 0$ ... $(i)$Let $M$ be the intersection of $OC$ and $AB$. $CM$ is the perpendicular distance from $C(3, 4)$ to the line $3x + 4y - 21 = 0$:$CM = \frac{|3(3) + 4(4) - 21|}{\sqrt{3^2 + 4^2}} = \frac{|9 + 16 - 21|}{\sqrt{25}} = \frac{4}{5}$In $	riangle AMC$,$\angle AMC = 90^\circ$. By Pythagoras theorem:$AM = \sqrt{AC^2 - CM^2} = \sqrt{2^2 - (\frac{4}{5})^2} = \sqrt{4 - \frac{16}{25}} = \sqrt{\frac{100 - 16}{25}} = \sqrt{\frac{84}{25}} = \frac{2\sqrt{21}}{5}$Since $AB = 2AM$,we have:$AB = 2 	imes \frac{2\sqrt{21}}{5} = \frac{4}{5}\sqrt{21}$

If $OA$ and $OB$ are the tangents to the circle $x^2 + y^2 - 6x - 8y + 21 = 0$ drawn from the origin $O$,then $AB =$

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