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.

Question

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.

Reiny · Answer

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.