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(A) The equation of the chord of contact from the origin $(0, 0)$ to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is $gx + fy + c = 0$.The equation of t

Vedclass · Accepted Answer

$\frac{1}{2} \left( \frac{g^2 + f^2 - c}{\sqrt{g^2 + f^2}} \right)$

Vedclass · Answer

(A) The equation of the chord of contact from the origin $(0, 0)$ to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is $gx + fy + c = 0$.The equation of the chord of contact from the point $(g, f)$ is $gx + fy + g(x + g) + f(y + f) + c = 0$,which simplifies to $2gx + 2fy + g^2 + f^2 + c = 0$,or $gx + fy + \frac{g^2 + f^2 + c}{2} = 0$.The distance $d$ between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.Here,$A = g$,$B = f$,$C_1 = c$,and $C_2 = \frac{g^2 + f^2 + c}{2}$.Thus,$d = \frac{|c - \frac{g^2 + f^2 + c}{2}|}{\sqrt{g^2 + f^2}} = \frac{|\frac{2c - g^2 - f^2 - c}{2}|}{\sqrt{g^2 + f^2}} = \frac{|\frac{c - g^2 - f^2}{2}|}{\sqrt{g^2 + f^2}} = \frac{1}{2} \left( \frac{g^2 + f^2 - c}{\sqrt{g^2 + f^2}} \right)$.

The distance between the chords of contact of the tangents to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ from the origin and the point $(g, f)$ is

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