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Two line segments $\mathrm{AB}$ and $\mathrm{CD}$ are constrained to move along the $\mathrm{x}$ and $\mathrm{y}$ axes, respectively, in such a way that the points A, B, C, D are concyclic. If $A B=a$ and $C D=b$, then the locus of the centre of the circle passing through A, B, C, D in polar coordinates is
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Verified Answer
The correct answer is:
$r^{2} \cos 2 \theta=\frac{a^{2}-b^{2}}{4}$
$$
\begin{array}{l}
2 \sqrt{g^{2}-c}=a \\
2 \sqrt{f^{2}-c}=b
\end{array}
$$
Polar coordinates of centre of circle be $(\cos \theta, r \sin \theta)$
$$
\begin{array}{ll}
g=-r \cos \theta \quad \text { and } g^{2}-f^{2}=\frac{a^{2}-b^{2}}{4} \\
f=-r \sin \theta \quad \therefore \quad r^{2} \cos 2 \theta=\frac{a^{2}-b^{2}}{4}
\end{array}
$$
\begin{array}{l}
2 \sqrt{g^{2}-c}=a \\
2 \sqrt{f^{2}-c}=b
\end{array}
$$
Polar coordinates of centre of circle be $(\cos \theta, r \sin \theta)$
$$
\begin{array}{ll}
g=-r \cos \theta \quad \text { and } g^{2}-f^{2}=\frac{a^{2}-b^{2}}{4} \\
f=-r \sin \theta \quad \therefore \quad r^{2} \cos 2 \theta=\frac{a^{2}-b^{2}}{4}
\end{array}
$$
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