Question 951654
Write the equation of the circle with center (4,-2) and containing the point (-4,13)
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Standard form of equation for a circle: (x-h)^2+(y-k)^2=r^2, r=radius, (h,k)=coordinates of center
Solving for r^2 using coordinates of given point(-4,13) on the circle
 (-4-4)^2+(13+2)^2=r^2
64+225=r^2
r^2=289
equation: (x-4)^2+(y+2)^2=289