Question

A circle passes throughthe origin and the point (5/2,1/2) and has 2y-3x=0 as a diameter. Find its equation.

Answer 1

(x-a)^2 + (y-b)^2 = r^2
(5/2-a)^2 + (1/2-b)^2 = r^2
25/4-5a+a^2+1/4-b+b^2 = r^2
but
a^2+b^2 = r^2 for (0,0)
so
26/4 -5a - b = 0
5 a + b = 26/4
center lies on 2y=3x or through(a,b)
2 b = 3 a
so
26/4 = 5 a + (3/2) a
26 = 20 a + 6 a = 26 a
a = 1
b = 3/2
take it from there

Answer 2

At any point (x, 3/2 x) on the line, the distance from the origin is

d^2 = x^2 + 9/4 x^2 = 13/4 x^2

If the circle's center is at (h,3h/2), then, we have

(x-h)^2 + (y-(3h/2))^2 = 13/4 h^2
(5/2-h)^2 + (1/2-(3h/2))^2 = 13/4 h^2
h=1

So, the circle is
(x-1)^2 + (y-3/2)^2 = 13/4

You can see from the plots below that the circle meets the requirements