a. The general equation for a circle centered at [tex](a,b)[/tex] with radius [tex]r[/tex] is[tex](x-a)^2+(y-b)^2=r^2[/tex]The described circle has equation[tex](x+3)^2+(y+2)^2=r^2[/tex]We know the circle passes through the origin. This means that the equation above holds for [tex]x=0[/tex] and [tex]y=0[/tex]. The distance between any point on the circle and its center is the radius, so we can use this fact to determine [tex]r[/tex]:[tex](0+3)^2+(0+2)^2=r^2\implies 9+4=13=r^2\implies r=\sqrt{13}[/tex]So the circle's equation is[tex](x+3)^2+(y+2)^2=(\sqrt{13})^2=13[/tex]b. If the distance between point B and the center is less than [tex]\sqrt{13}[/tex], then B lies inside the circle. If the distance is greater than [tex]\sqrt{13}[/tex], it falls outside the circle. Otherwise, if the distance is exactly [tex]\sqrt{13}[/tex], then B lies on the circle.The distance from B to the center is[tex]\sqrt{(-1+3)^2+(3+2)^2}=\sqrt{4+25}=\sqrt{29}[/tex][tex]29>13[/tex], so [tex]\sqrt{29}>\sqrt{13}[/tex], which means B falls outside the circle.