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(B) Let the equation of the circle passing through $A, B, C, D$ be $x^2+y^2+2gx+2fy+c=0$.The polar coordinates of the centre $(-g, -f)$ are given by $

Vedclass · Accepted Answer

$r^2 \cos 2	heta = \frac{a^2-b^2}{4}$

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(B) Let the equation of the circle passing through $A, B, C, D$ be $x^2+y^2+2gx+2fy+c=0$.The polar coordinates of the centre $(-g, -f)$ are given by $-g = r \cos 	heta$ and $-f = r \sin 	heta$.$AB$ is the length of the intercept on the $X$-axis,so $a = 2\sqrt{g^2-c} \implies \frac{a^2}{4} = g^2-c$.$CD$ is the length of the intercept on the $Y$-axis,so $b = 2\sqrt{f^2-c} \implies \frac{b^2}{4} = f^2-c$.Subtracting the two equations: $g^2-f^2 = \frac{a^2-b^2}{4}$.Substituting $g = -r \cos 	heta$ and $f = -r \sin 	heta$:$(-r \cos 	heta)^2 - (-r \sin 	heta)^2 = \frac{a^2-b^2}{4}$$r^2(\cos^2 	heta - \sin^2 	heta) = \frac{a^2-b^2}{4}$$r^2 \cos 2	heta = \frac{a^2-b^2}{4}$.

Two line segments $AB$ and $CD$ are constrained to move along the $X$ and $Y$-axes,respectively,in such a way that the points $A, B, C, D$ are concyclic. If $AB = a$ and $CD = b$,then the locus of the centre of the circle passing through $A, B, C, D$ in polar coordinates is

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