Two rods of lengths a and b slide along the c | vedclass

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Question

(A) Let the circle passing through the ends of the rods be $x^2 + y^2 + 2gx + 2fy + c = 0$.The rod of length $a$ lies on the $x$-axis,so its intercept

Vedclass · Accepted Answer

$4(x^2 - y^2) = a^2 - b^2$

Vedclass · Answer

(A) Let the circle passing through the ends of the rods be $x^2 + y^2 + 2gx + 2fy + c = 0$.The rod of length $a$ lies on the $x$-axis,so its intercepts on the $x$-axis are the roots of $x^2 + 2gx + c = 0$. The length of the intercept is $2\sqrt{g^2 - c} = a$,which implies $4(g^2 - c) = a^2$.The rod of length $b$ lies on the $y$-axis,so its intercepts on the $y$-axis are the roots of $y^2 + 2fy + c = 0$. The length of the intercept is $2\sqrt{f^2 - c} = b$,which implies $4(f^2 - c) = b^2$.Subtracting the two equations: $4(g^2 - f^2) = a^2 - b^2$.Since the center of the circle is $(h, k) = (-g, -f)$,we have $g = -h$ and $f = -k$.Substituting these into the equation: $4((-h)^2 - (-k)^2) = a^2 - b^2$,which simplifies to $4(h^2 - k^2) = a^2 - b^2$.Thus,the locus of the center $(h, k)$ is $4(x^2 - y^2) = a^2 - b^2$.

Two rods of lengths $a$ and $b$ slide along the coordinate axes in such a way that their ends are always concyclic. The locus of the center of the circle passing through the ends is:

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