## Thomas' Calculus 13th Edition

$(x+\sqrt{3})^{2}+(y+2)^{2}=4$ .
The circle with center $C(h,k)$ and radus $a$ has the general equation $( x-h)^{2}+(y-k)^{2}=a^{2}.$ Here, $\qquad$ $( x-(-\sqrt{3}))^{2}+(y-(-2))^{2}=2^{2}$ $(x+\sqrt{3})^{2}+(y+2)^{2}=4$ x-intercepts (y=0): $(x+\sqrt{3})^{2}+(0+2)^{2}=4$ $(x+\sqrt{3})^{2}+4=4$ $(x+\sqrt{3})^{2}=0$ $x=-\sqrt{3}\qquad$... $\quad (-\sqrt{3},0)$ y-intercepts (x=0): $(0+\sqrt{3})^{2}+(y+2)^{2}=4$ $3+(y+2)^{2}=4$ $(y+2)^{2}=1$ $y+2=\pm 1$ $y=-2\pm 1$ $y=-1,\quad y=-3\qquad (0,-1),(0,-3)$