How to find the equation of a circle given 3 points

Question

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)

Answer 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

$(x-2)^2+y^2=(x-4)^2+(y+2)^2$;

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 $r^2$.

Answer 2

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

Answer 3

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