Find the distance between the chords of conta | vedclass

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(D) The equation of the chord of contact for a point $(x_1, y_1)$ with respect to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is given by $xx_1 + yy_1

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$\frac{g^2 + f^2 - c}{2 \sqrt{g^2 + f^2}}$

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(D) The equation of the chord of contact for a point $(x_1, y_1)$ with respect to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is given by $xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0$.For the point $(0, 0)$,the chord of contact is $g(x + 0) + f(y + 0) + c = 0$,which simplifies to $gx + fy + c = 0$ (Equation $1$).For the point $(g, f)$,the chord of contact is $xg + yf + g(x + g) + f(y + f) + c = 0$,which simplifies to $2gx + 2fy + g^2 + f^2 + c = 0$,or $gx + fy + \frac{1}{2}(g^2 + f^2 + c) = 0$ (Equation $2$).The distance 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 $C_2 = \frac{1}{2}(g^2 + f^2 + c)$.Distance $d = \frac{|\frac{1}{2}(g^2 + f^2 + c) - c|}{\sqrt{g^2 + f^2}} = \frac{|\frac{1}{2}(g^2 + f^2 - c)|}{\sqrt{g^2 + f^2}} = \frac{g^2 + f^2 - c}{2 \sqrt{g^2 + f^2}}$.

Find the distance between the chords of contact of the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ with respect to the points $(0, 0)$ and $(g, f)$.

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