The area of the triangle formed by the tangen | vedclass

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Question

(A) Let the point be $P(h, k)$ and the circle be $x^2 + y^2 = a^2$. The equation of the chord of contact $AB$ is $xh + yk = a^2$. The length of the pe

Vedclass · Accepted Answer

$a \frac{(h^2 + k^2 - a^2)^{3/2}}{h^2 + k^2}$

Vedclass · Answer

(A) Let the point be $P(h, k)$ and the circle be $x^2 + y^2 = a^2$. The equation of the chord of contact $AB$ is $xh + yk = a^2$. The length of the perpendicular $OM$ from the center $O(0, 0)$ to the chord $AB$ is $OM = \frac{a^2}{\sqrt{h^2 + k^2}}$. In the right-angled triangle $OAM$,$AM = \sqrt{OA^2 - OM^2} = \sqrt{a^2 - \frac{a^4}{h^2 + k^2}} = \frac{a\sqrt{h^2 + k^2 - a^2}}{\sqrt{h^2 + k^2}}$. Since $AB = 2AM$,we have $AB = \frac{2a\sqrt{h^2 + k^2 - a^2}}{\sqrt{h^2 + k^2}}$. The length of the perpendicular $PM$ from $P(h, k)$ to the chord $AB$ is $PM = OP - OM = \sqrt{h^2 + k^2} - \frac{a^2}{\sqrt{h^2 + k^2}} = \frac{h^2 + k^2 - a^2}{\sqrt{h^2 + k^2}}$. The area of triangle $PAB$ is $\frac{1}{2} 	imes AB 	imes PM = \frac{1}{2} 	imes \left( \frac{2a\sqrt{h^2 + k^2 - a^2}}{\sqrt{h^2 + k^2}} ight) 	imes \left( \frac{h^2 + k^2 - a^2}{\sqrt{h^2 + k^2}} ight) = \frac{a(h^2 + k^2 - a^2)^{3/2}}{h^2 + k^2}$.

The area of the triangle formed by the tangents from the point $(h, k)$ to the circle $x^2 + y^2 = a^2$ and the line joining their points of contact is

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