How to find the equation of a circle given 3 points Sorry in advance for the long post; I am doing a set of questions on finding the equation of a circle given 3 points.
The first set of points are (0,0), (2,0), and (4,-2).
There are a few ways I can see of doing this question but I can't help feeling there is something I am meant to spot that I haven't yet - possibly to do with circle theorems.
The obvious idea is to create 3 simultaneous equations and solve them for a, b, and r, where (a,b) is the centre of the circle, and r is the radius.
But this seems really laborious.
The second idea I had is to somehow use the circle theorem that the angle subtended at the circumference is half that at the centre, but I couldn't see an obvious way in.
The way I ended up solving it was by considering the symmetry of a circle;
If (0,0) and (2,0) form a chord of the circle which is parallel to the x-axis, then the centre of the circle must have an x-coordinate in line with the midpoint of the chord.
So the x-coordinate of the centre is 1.
Then to continue the question, if you consider symmetry again, there must be a coordinate at (-2,-2).
And then just by looking at these coordinates, by symmetry again there is another coordinate at (-2, -4) and (4, -4). At this point, looking at the sketch (with the help of Desmos), you can see that the centre's y-coordinate is at -3.
This second bit of reasoning to find the y-coordinate is a bit unsatisfactory and I can't fully explain it.
Can anyone suggest a neater way of going about this problem?
And/or explain why my method was good or bad?
P.S. The rest of the set of questions were:


*

*(2,2), (4,3), & (6,9)

*(-1,1), (2,-1), & (-2,0)

*(0,0), (a,0, & (1,1)

 A: The center of the circle containing $(0,0), (2,0), $ and $(4,-2)$ 
must be equidistant from all those points.
The points equidistant from $(0,0)$ and $(2,0)$ are on a line satisfying $x^2+y^2=(x-2)^2+y^2$; 
i.e., $0=-4x+4$; i.e., $x=1$.
The points equidistant from $(2,0)$ and $(4,-2)$ are on a line satisfying 
$(x-2)^2+y^2=(x-4)^2+(y+2)^2$;
i.e., $-4x+4=-8x+16+4y+4$; i.e., $4x=4y+16$; i.e., $y=x-4$.
Therefore, the center is at the intersection of these lines, which is $(1,-3)$.
Therefore, the equation of the circle is $(x-1)^2+(y+3)^2=r^2$.
Plug in any of the points to figure out $r^2$.
A: Take two points and find the perpendicular bisector. Repeat for another pair of points. The centre of the circle is where these perpendicular bisectors meet
A: Use family of circles i.e circle with $(0,0)$ and $(2,0)$ as one of its diameter is $$x(x-2)+y^2=0$$ 
And line passing through these points is $y=0$ Hence any circle passing through the point of intersection of circle and line is $$x^2+y^2-2x+\lambda y=0$$ 
Required circle passes through the point $(4,-2)$ satisfy that we will get parameter $\lambda$.
$$ 4^2+(-2)^2-2(4)+\lambda (-2)=0$$
$$\therefore \lambda =6$$
Hence equation of required circle is $x^2+y^2-2x+6y=0$
