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(A) The given equation of the pair of lines is $ax^2+2hxy-ay^2+2gx+2fy+c=0$. Let the point of intersection be $(x_0, y_0)$. The coordinates of the poi

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$\frac{f^2+g^2}{a^2+h^2}$

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(A) The given equation of the pair of lines is $ax^2+2hxy-ay^2+2gx+2fy+c=0$. Let the point of intersection be $(x_0, y_0)$. The coordinates of the point of intersection of the pair of lines $Ax^2+2Hxy+By^2+2Gx+2Fy+C=0$ are given by the formulas: $x_0 = \frac{HF-BG}{AB-H^2}$ and $y_0 = \frac{GH-AF}{AB-H^2}$. Here,$A=a$,$H=h$,$B=-a$,$G=g$,$F=f$,$C=c$. Substituting these values: $x_0 = \frac{hf-g(-a)}{a(-a)-h^2} = \frac{hf+ag}{-(a^2+h^2)}$ $y_0 = \frac{gh-af}{-(a^2+h^2)} = \frac{af-gh}{a^2+h^2}$. The square of the distance from the origin $(0,0)$ is $x_0^2 + y_0^2$. $x_0^2 + y_0^2 = \frac{(hf+ag)^2 + (af-gh)^2}{(a^2+h^2)^2} = \frac{h^2f^2+a^2g^2+2afgh + a^2f^2+g^2h^2-2afgh}{(a^2+h^2)^2}$ $= \frac{f^2(h^2+a^2) + g^2(a^2+h^2)}{(a^2+h^2)^2} = \frac{(f^2+g^2)(a^2+h^2)}{(a^2+h^2)^2} = \frac{f^2+g^2}{a^2+h^2}$.

The square of the distance from the origin to the point of intersection of the pair of lines $ax^2+2hxy-ay^2+2gx+2fy+c=0$ is

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