Tangents are drawn from the point (4, 3) to t | vedclass

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Question

(C) Let the point be $P(4, 3)$ and the circle be $x^2 + y^2 = 9$. The radius $r = 3$.The equation of the chord of contact $AB$ is $T = 0$,which is $4x

Vedclass · Accepted Answer

$\frac{192}{25}$

Vedclass · Answer

(C) Let the point be $P(4, 3)$ and the circle be $x^2 + y^2 = 9$. The radius $r = 3$.The equation of the chord of contact $AB$ is $T = 0$,which is $4x + 3y = 9$.The distance from the origin $O(0, 0)$ to the chord $AB$ is $OQ = \frac{|4(0) + 3(0) - 9|}{\sqrt{4^2 + 3^2}} = \frac{9}{5}$.In $	riangle OAQ$,$AQ = \sqrt{OA^2 - OQ^2} = \sqrt{3^2 - (\frac{9}{5})^2} = \sqrt{9 - \frac{81}{25}} = \sqrt{\frac{225 - 81}{25}} = \sqrt{\frac{144}{25}} = \frac{12}{5}$.The length of the chord of contact $AB = 2 	imes AQ = 2 	imes \frac{12}{5} = \frac{24}{5}$.The distance $PQ$ is the distance from $P(4, 3)$ to the line $4x + 3y - 9 = 0$,which is $PQ = \frac{|4(4) + 3(3) - 9|}{\sqrt{4^2 + 3^2}} = \frac{|16 + 9 - 9|}{5} = \frac{16}{5}$.The area of $	riangle PAB = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes AB 	imes PQ = \frac{1}{2} 	imes \frac{24}{5} 	imes \frac{16}{5} = \frac{192}{25}$.

Tangents are drawn from the point $(4, 3)$ to the circle $x^2 + y^2 = 9$. The area of the triangle formed by them and the line joining their points of contact is

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