Question 61514
Find the standard equation of the circle that satisfies the conditions.

Center (-1,-3), passing through the point (3,0)

In order to write an equation for a circle you need a center and a radius.  You're missing a radius, but you can find it with the information that they gave you.  The radius is the distance from the center to any point of the circle.
The distance formula is r={{{highlight(d=sqrt((x2-x1)^2+(y2-y1)^2))}}}
(x1,y1)=(-1,3) and (x2,y2)=(3,0)
r={{{d=sqrt((3-(-1))^2+(0-3)^2)}}}
r={{{d=sqrt((4)^2+(-3)^2)}}}
r={{{d=sqrt(16+9)}}}
r={{{d=sqrt(25)}}}
r={{{d=5}}}
Now that we have a center (h,k)=(-1,-3) and a radius r=5.
The standard form of a circle:  {{{(x-h)^2+(y-k)^2=r^2}}}, (h,k) is the center, and r is the radius.
{{{(x-(-1))^2+(y-(-3))^2=5^2}}}
{{{highlight((x+1)^2+(y+3)^2=25)}}}
:
Happy Calculating!!!