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A circle of radius $\sqrt3-1$ units with both coordinates of the centre negative, touches the straight lines $y-\sqrt3x=0$ and $x-\sqrt3y=0$. Prove that the equation of the circle is $$ x^2+y^2+4(x+y)+(\sqrt3+1)^2=0. $$

My Approach

Let $(x_1,y_1)$ be the coordinates of the centre of the circle. Since the circle is in the third quadrant and touches both $y-\sqrt3x=0$ and $x-\sqrt3y=0$, we can say tangents to the above two lines meet at the center $(x_1,y_1)$. In that case, \begin{equation} \tag{1} y-y_1=-\sqrt3(x-x_1) \end{equation} and \begin{equation} \tag{2} y-y_1 = - \dfrac{1}{\sqrt3} (x - x_1) \end{equation}

Also using the point to line distance formula we get \begin{equation} \tag{3} \dfrac{|y_1-\dfrac{1}{\sqrt3}x_1|}{\sqrt{1+\dfrac{1}{3}}} =\sqrt3-1 \end{equation}

From (3) we get $y_1-\dfrac{1}{\sqrt3}x_1=2-\dfrac{2}{\sqrt3}$

Also using the same line to distance formula for the other line we get,

\begin{equation} \tag{4} y_1-\sqrt3 x_1=2(\sqrt3-1) \end{equation}

From (3) and (4) we get values of $x_1=2\sqrt3-4$ and similarly for $y_1$.

My question

Apart from that is there any shorter method to obtain the center coordinates and subsequently the equation of the given circle?

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We can see that the center $C$ of the circle must lie on angle bisector of these two lines $y=x$ (since their angles wit x-axsis respectively $60^{\circ}$ and $30^{\circ}$. So $C = (-p,-p)$ for some positive $p$. Now $CO = p\sqrt{2}$ (where $O$ is origin) so we have $$\sin 15^{\circ} ={\sqrt{3}-1\over p\sqrt{2}}\implies p=2$$

So the circle has equation $$(x+2)^2+(y+2)^2=(\sqrt{3}-1)^2$$

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