if you have endpoints of a diameter (0,0) and (6,8) and you have to write the standard form of the equation of a circle, what would the origin be? i did the question assuming the origin is (0,0) and got the equation x^2 +y^2=25. but apparently thats wrong and the equation is (x-3)^2 + (y-4)^2 =25.. meaning that the origin is assumed to be (3,4).... basically how would one know what the origin is in this instance as both equations are diff?

Question

if you have endpoints of a diameter (0,0) and (6,8) and you have to write the standard form of the equation of a circle, what would the origin be?  i did the question assuming the origin is (0,0) and got the equation x^2 +y^2=25. but apparently thats wrong and the equation is (x-3)^2 + (y-4)^2 =25.. meaning that the origin is assumed to be (3,4).... basically how would one know what the origin is in this instance as both equations are diff?

anonymous · Answer

by "origin" do you mean "center"?

anonymous · Answer

if (0,0) is one endpoint of the diameter it is not also the center.
since the other endpoint is (6,8) the center is half way between 
i.e. at the point (3,4)

anonymous · Answer

yes i mean center of the circle

anonymous · Answer

omg okay thanks blond moment :)

anonymous · Answer

so your equation will look like 
$$(x-3)^2+(y-4)^2=r^2$$ and you find $r^2$ by pythagoras

anonymous · Answer

square of the distance between (0,0) and (3,4) is $3^2+4^2=25$
gentlemen prefer blondes

anonymous · Answer

haha yeah yeah. thanks for the help :)