Thread: Coordinate Geometry - equation of the circle

I don't know how to start solving this problem.
I have drawn the graph but have no idea to start.

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^(1/2)).
Find the equation of the circle C.

Can anyone give me some hints?
Thank in advance.

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!

I actually had the same approach as Soroban.
---
Let me add something.
It seems to me there are 2 possible circles here.
We are told that the center of the circle is located sqrt(1/5)
from the line x-2y=0.
That means the locus of point satisfing the condition that it needs to be located sqrt(1/5) is a pair of parallel line whose distance is sqrt(1/5) from x-2y=0.
Since the line is parallel its equation is,
y=(1/2)x+b
One point on the line is: (0,b)
The distance from x-2y=0 is sqrt(1/5)
Thus, by distance formula for a line,
|-2b+0|/sqrt(1+2^2)=sqrt(1/5)
Thus,
|-2b|=1
Thus,
b=+/- 1/2
Thus, there are two possible lines on which the center of circle lies upon.