Find equation of circle with radius $\sqrt3-1$ units with both coordinates of the centre negative

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?

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 $$(x+2)^2+(y+2)^2=(\sqrt{3}-1)^2$$