<PRE><B>Question 1119612</B>
For the last one
(x-a)^2+(y-b)^2=r^2
For the origin, x and y are 0 so a^2+b^2=r^2
For (6, 0) (6-a)^2+(0-b^2)=r^2 and 36-12a+a^2+b^2=r^2
But since r^2=a^2+b^2
 36-12a+a^2+b^2=a^2+b^2
12a=36 and a=3
similarly, for (0, -8) we have a^2+64+16b+b^2=a^2+b^2 and -16b=-64 and b=-4
The center is at (3, -4) and the radius is sqrt((-3)^2+4^2)=5
(x-3)^2+(y+4)^2=25

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