Let a chord AB subtend an angle of 60^{circ} | vedclass

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(C) The length of the perpendicular $CD$ from the centre $C(2,3)$ to the chord $AB$ $(x+y+1=0)$ is given by:$CD = \left|\frac{2+3+1}{\sqrt{1^2+1^2}}\r

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$x^2+y^2-4x-6y-11=0$

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(C) The length of the perpendicular $CD$ from the centre $C(2,3)$ to the chord $AB$ $(x+y+1=0)$ is given by:$CD = \left|\frac{2+3+1}{\sqrt{1^2+1^2}}ight| = \frac{6}{\sqrt{2}} = 3\sqrt{2}$.In $	riangle CAD$,the angle at the centre is $60^{\circ}$,so $\angle ACD = 30^{\circ}$.Using trigonometry in $	riangle CAD$:$\cos 30^{\circ} = \frac{CD}{AC} \Rightarrow \frac{\sqrt{3}}{2} = \frac{3\sqrt{2}}{AC}$.$AC = \frac{6\sqrt{2}}{\sqrt{3}} = 2\sqrt{6}$.The radius $r = AC$,so $r^2 = (2\sqrt{6})^2 = 4 	imes 6 = 24$.The equation of the circle with centre $(2,3)$ and radius squared $24$ is:$(x-2)^2 + (y-3)^2 = 24$$x^2 - 4x + 4 + y^2 - 6y + 9 = 24$$x^2 + y^2 - 4x - 6y + 13 = 24$$x^2 + y^2 - 4x - 6y - 11 = 0$.

Let a chord $AB$ subtend an angle of $60^{\circ}$ at the centre $C(2,3)$ of a circle $S$. If the equation of $AB$ is $x+y+1=0$,then the equation of the circle $S$ is

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