Asked by Therese
I had to write the standard form of the equation for the circle consisting of the endpoints (0,0), (6,8). I got (x-3)^2+(y-4)^2=15. But my book said the fifteen should be a twenty five. How would they have gotten 25? Thank you.
Answers
Answered by
Reiny
looks like you got the centre correct
let the equation be
(x-3^2 + (y-4)^2 = r^2
but (0,0) lies on it, so
(-3)^2 + (-4)^2 = r^2
9 + 16 = R^2
=r^2 = 25
then
(-3)^2 + (-4)^2 = 25
What I don't see is how you got 15.
let the equation be
(x-3^2 + (y-4)^2 = r^2
but (0,0) lies on it, so
(-3)^2 + (-4)^2 = r^2
9 + 16 = R^2
=r^2 = 25
then
(-3)^2 + (-4)^2 = 25
What I don't see is how you got 15.
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