A circle passes through origin and the point (5/2, 1/2) and has 2y_3x=0 as a diameter. Find its equation.

Answer 1

The center lies on the perpendicular bisector of the chord joining (0,0) and (5/2,1/2). That chord has
slope = 1/5
midpoint = (5/4,1/4)
So, the bisector is the line

y - 1/4 = -5(x - 5/4)

The center of the circle is at the intersection of the bisector and the line 2y=3x.

So, the center is at (1,3/2)

Since it passes through (0,0), r^2 = 13

That means the circle is

(x-1)^2 + (y-3/2)^2 = 13/4
see the graph at
http://www.wolframalpha.com/input/?i=plot+(x-1)%5E2+%2B+(y-3%2F2)%5E2+%3D+13%2F4,+2y%3D3x,x%3D5%2F2,y%3D1%2F2

or, you know that

(0-h)^2 + (0-k)^2 = r^2
(5/2-h)^2 + (1/2-k)^2 = r^2
2k=3h

solve those and you get the three values.