The locus of a point,such that the difference | vedclass

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(C) Let the two circles be $S_1 = 0$ and $S_2 = 0$. Let the point be $P(x, y)$. The square of the length of the tangent from $P$ to $S_1 = 0$ is $t_1^

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Straight line

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(C) Let the two circles be $S_1 = 0$ and $S_2 = 0$. Let the point be $P(x, y)$. The square of the length of the tangent from $P$ to $S_1 = 0$ is $t_1^2 = S_1$. The square of the length of the tangent from $P$ to $S_2 = 0$ is $t_2^2 = S_2$. Given that $t_1^2 - t_2^2 = k$,where $k$ is a constant. Substituting the expressions,we get $S_1 - S_2 = k$. Since $S_1$ and $S_2$ are second-degree equations in $x$ and $y$ of the form $x^2 + y^2 + 2g_ix + 2f_iy + c_i = 0$,the difference $S_1 - S_2$ results in a linear equation of the form $2(g_1 - g_2)x + 2(f_1 - f_2)y + (c_1 - c_2) = k$. This is a first-degree equation in $x$ and $y$,which represents a straight line parallel to the radical axis $S_1 - S_2 = 0$.

The locus of a point,such that the difference of the squares of the lengths of the tangents drawn from it to two given circles is constant,is:

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