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Let a chord $\mathrm{AB}$ subtend at angle of $60^{\circ}$ at the centre $\mathrm{C}(2,3)$ of a circle $\mathrm{S}$. If the equation of $\mathrm{AB}$ is $\mathrm{x}+\mathrm{y}+1=0$, then the equation of the circle $\mathrm{S}$ is
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The correct answer is:
$x^2+y^2-4 x-6 y-11=0$
Length of perpendicular from $C$ on $A B$
$$
C D=\left|\frac{2+3+1}{\sqrt{1+1}}\right|=\frac{6}{\sqrt{2}}=3 \sqrt{2}
$$
In $\triangle C A D$,
$$
\sin 60^{\circ}=\frac{C D}{A C} \Rightarrow \frac{\sqrt{3}}{2}=\frac{3 \sqrt{2}}{A C} \Rightarrow A C=\frac{6 \sqrt{2}}{\sqrt{3}}
$$
$\therefore$ Equation of $S:(x-2)^2+(y-3)^2=\frac{36 \times 2}{3}$
$$
\begin{aligned}
& \Rightarrow x^2+y^2-4 x-6 y+13=24 \\
& \Rightarrow x^2+y^2-4 x-6 y-11=0 .
\end{aligned}
$$
$$
C D=\left|\frac{2+3+1}{\sqrt{1+1}}\right|=\frac{6}{\sqrt{2}}=3 \sqrt{2}
$$
In $\triangle C A D$,
$$
\sin 60^{\circ}=\frac{C D}{A C} \Rightarrow \frac{\sqrt{3}}{2}=\frac{3 \sqrt{2}}{A C} \Rightarrow A C=\frac{6 \sqrt{2}}{\sqrt{3}}
$$
$\therefore$ Equation of $S:(x-2)^2+(y-3)^2=\frac{36 \times 2}{3}$
$$
\begin{aligned}
& \Rightarrow x^2+y^2-4 x-6 y+13=24 \\
& \Rightarrow x^2+y^2-4 x-6 y-11=0 .
\end{aligned}
$$
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