SOLUTION: {{{highlight(A_circle_C)}}} x^2+y^2-2gx+2fy+e=0 has its center in the first quadrant. The x and y axes are tangents to C and {{{highlight(cross(its))}}} the point (3,6) is on the

1.  Since the axes x- and y- are tangents to C, the center of C lies on the line x = y. (which is the diagonal of the QI).

    In other words, the center of the circle is at the point (x0,x0) for some x0 (now unknown).

2.  Due to the same reason, the radius of the circle is equal to x0 (and to y0, too).

3.  So, the equation of the circle is

     = .   (1)

    The point (3,6) lies on the circle, so coordinates satisfy the equation (1). In other words,

     = .

4.  It gives you a quadratic equation for .

     = .

    Simplify and solve for :

     = .

    The roots are   = 5  and   = 9.

5.  The first circle has the center (5,5) and the radius 5.

    Its equation is  = 25.


    The second circle has the center (9,9) and the radius 9.

    Its equation is  = 81.