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)

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


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 $r^2$.


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


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 $x^2+y^2-2x+6y=0$


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