SOLUTION: "Write the equation of the circle with center (4,-2) and containing the point (-4,13)"

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Question 951654: "Write the equation of the circle with center (4,-2) and containing the point (-4,13)"
Found 2 solutions by josgarithmetic, lwsshak3:
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
The radius is the distance from center (4,-2) to the given point (-4,13) on the circle. Use the distance formula to find this.

In raw form according to Standard Form equation for a circle,
;
Simplify this and you have your needed result.

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Write the equation of the circle with center (4,-2) and containing the point (-4,13)
***
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