I am trying to solve this problem:
We know that there's a circle with center$(m,h)$. And it passes through the points(1,0), (-1,0).
Show that there's a unique circle passing through the three points:$(1,0),(-1,0),(x_0,y_0)$.
I tried making substitution, and get:
$$h=\frac{{x_0}^2+{y_0}^2-1}{2{y_0}}$$
I said that only 1 value of h would be produced, so there's a unique value of h implying that a unique circle would pass the point.
After checking the answer, it said that:
Given:
$${x_0}^2+{y_0}^2=1+2y_0\sqrt{r^2-1}(1)$$
I know how the (1) is deduced, but I don't know why does it show that there's unique circle satisfying the condition. The answer justified this by saying that the equation is uniquely dependent on $x_0, y_0$. I don't know what does this mean.
Thank you very much for you guys.
Note: r is the radius of the circle.
The fact that the circle passes through $A=(1,0)$ and $B=(-1,0)$ is equivalent to the fact that its center $C$ (which is necessarily on the line bissector of $AB$) has coordinates $(0,a)$.
Therefore, the circle being the locus of points $M$ such that
$$CM^2=CA^2,$$
if we take $M=(x,y)$, this relationship is converted into the equation
$$x^2+(y-a)^2=1+a^2$$
or (simplifying by $a^2$)
$$x^2+y^2-2ay-1=0\tag{1}$$
Saying that this circle passes through $(x_0,y_0)$ is equivalent to say that :
$$x_0^2+y_0^2-2ay_0-1=0\tag{2}$$
One obtains a first degree equation in $a$, explaining that there exists a unique solution,
$$a=\dfrac{x_0^2+y_0^2-1}{2y_0}$$
(the equation you mention) with an exceptional case : no solution exists when the denominator $y_0=0$, i.e., when $M_0$ belongs to the $x$-axis, which is natural because no circle passes through 3 aligned points.