Hello, ling_c_0202!

I have an approach, but it's very messy.

. . I'll leave the algebra up to you . . .

A circle C cuts the x-axis at A and B, and the y-axis at P and Q.

It is given that AB = 2 and (major arc PQ):(minor arc PQ = 3:1). . . . . . . . . ._

The distance from the centre G of the circle C to the line x - 2y = 0 is d = 1/√5

Find the equation of the circle C. Code:

| * * *
| * *
Q* *
*| * *
| *
* | * G *
* | * *
* | ** * *
| * * *
P*| * * * *
* * * *
----+--*---------*------
| A * * * B

The center of the circle is: G(h,k); its radius is *r.*

Let the four intercepts be: .A(a,0), B(b,0), P(0,p), Q(0.q)

We have seven unknowns to solve for

. . but we have enough equations.

GA = r: . (h - a)² + k² .= .r²

BG = r: . (h - b)² + k² .= .r²

GP = 4: . h² + (k - p)² .= .r²

GQ = r: . h² + (k - q)² .= .r²

We know that: .b - a .= .2

From the ratio of the arcs, we know that: angle PGQ = 90°.

Then: .PQ] .= .r√2 . . . Hence: .q .= .p + r√2

Since GP is perpendicular to GQ, their slopes are negative reciprocals.

So that: .m(GP) = (k - p)/h is the negative reciprocal of m(GQ) = (k - q)/h

I *told* you it was messy . . . Good luck!