If OA and OB are tangents from the origin to | 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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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 $(0, 0)$ is given by $T = 0$,which is $x(0) + y(0) - 3(x + 0) - 4(y + 0) + 21 = 0$,simplifying to $3x + 4y - 21 = 0$.The perpendicular distance $CM$ from the center $C(3, 4)$ to the line $3x + 4y - 21 = 0$ is $CM = \frac{|3(3) + 4(4) - 21|}{\sqrt{3^{2} + 4^{2}}} = \frac{|9 + 16 - 21|}{5} = \frac{4}{5}$.In the right-angled triangle $	riangle AMC$,$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 tangents from the origin to the circle $x^{2} + y^{2} - 6x - 8y + 21 = 0$,then $AB = \dots$

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