Direct method to find the equation of a circle. Suppose we are given four concyclic points or two lines which intersect the axes in concyclic points. Many a times, one point has a variable as a co-ordinate. Suppose the concyclic points are $(a_1,0),\;(a_2,0),\;(0,b_1),\;(0,b_2)$. Then, the relation between them is $$a_1a_2=b_1b_2$$
(Though I don't know how it can be proved.)
Now, knowing the four points, is there any direct way to find the equation of the circle? Or do we have to find the '$g$', the '$f$', and the '$c$' of the equation $x^2+y^2+2gx+2fy+c=0$ using those points?
 A: You need to find the center of the circle. It can be found by averaging the x-intercepts and averaging the y-intercepts. That will give you the center (h,k).
    To average the x-intercepts:  ((a₁+a₂)/2)=h.
    To average the y-intercepts:  ((b₁+b₂)/2)=k.
    So, the center is at (h,k).
I was thinking that you knew the four points. If you don't know them, maybe you can get some ideas from this.
Then, use the general formula for a circle:
    (x-h)²+(y-k)²=r²
To find the radius, I believe you can just substitute the values of a point on the circle (a₁, 0) in for x and y to get the radius squared. Then write the equation substituting the values you got for h,k, and r. This works if you know the four points. Maybe there is something here that would help you if you don't know the four points.
A: Note: You can't pass a circle through any arbitrary four points. However, the formula for a circle to go through any three points is fairly simple. The condition for the fourth point to be on this circle is also simple.
If you do not have the formula for circle through three points, here is the method. If the center is at $(c_x, c_y)$ and radius is $R$
$$
(P_{1x}-c_x)^2 + (P_{1y}-c_y)^2 = R^2 \tag 1
$$
$$
(P_{2x}-c_x)^2 + (P_{2y}-c_y)^2 = R^2 \tag 2
$$
$$
(P_{3x}-c_x)^2 + (P_{3y}-c_y)^2 = R^2 \tag 3
$$
Subtract (1) from (2) to get
$$
P_{2x}^2 - P_{1x} + 2 (P_{1x}-P_{2x} )c_x 
P_{2y}^2 - P_{1y} + 2 (P_{1y}-P_{2y} )c_y = 0 \tag 4$$ 
Similarly (3)-(1) gives
$$
P_{3x}^2 - P_{1x} + 2 (P_{1x}-P_{3x})c_x 
P_{3y}^2 - P_{1y} + 2 (P_{1y}-P_{3y}) c_y = 0 \tag 5$$ 
(4) and (5) are linear and you can solve for $c_x$ and $c_y$ (Here you will see that for a solution to exist, the three points can't be collinear).
Once you have $c_x$ and $c_y$ you can solve for $R$ from any of the equation (1),(2), (3).
You can then see the condition for the 4th point to be on the cirlce.
BTW, if the three points are $(a_1,0)$, $(a_2,0)$, $(0, b_1)$, then the equation (4) and (5) are very simple and it is easy to see the condition for $(0,y)$ to be on the circle.
A: The fact that $a_1a_2 = b_1b_2$ follows from the fact that these are the lengths of segments of intersecting chords in the circle.  The fact that two cross the $x$ axis and two cross the $y$ axis means that the origin is contained.  This also means that $a_1$ has the opposite sign of $a_2$, and $b_1$ the opposite sign of $b_2$ (assuming all are $\neq 0$).
A: Let $O$ be the origin and let 
$A \equiv (a_1,0)$
$B \equiv (a_2,0)$
$C \equiv (0,b_1)$
$D \equiv (0,b_2)$
So now by power of point theorem on point $O$ We have $OA.OB=OC.OD$
we know that $OA=a_1,OB=a_2,OC=b_1,OD=b_2$
So $a_1.a_2=b_1.b_2$
