# Constructing Circles

Through a point in the plane, infinitely many lines can pass. However, through two distinct points in the plane, *exactly one* line can pass. That is, *two distinct points uniquely determine a line*.

What happens in the case of circles? How many points are required to uniquely determine a circle?

It should be obvious that through one point, infinitely many circles can pass. Even through two points, infinitely many circles can pass, as the following figure shows:

However, *exactly one circle will pass through three distinct points*. In other words, *three distinct points will uniquely determine a circle*. There is an important point though: the three distinct points must be *non-collinear*. If three distinct points are collinear, a circle cannot pass through them (can you visualize why?).

Let us now see how we can actually construct a circle passing through three distinct non-collinear points. The following figure shows three such points, A,B and C:

What will be the characteristic(s) of any circle passing through these three points? Well, the center of that circle must be equidistant from all the three points. This means that we have to locate *that* point in the plane (call it O) such that OA = OB = OC.

To locate O, recall that any point which is equidistant from two fixed points *must lie on the perpendicular bisector* of the segment joining those points. Therefore, we proceed as follows:

1. First, we draw the perpendicular bisector P_{1} of AB. Any point on P_{1} will be equidistant from A and B:

2. Next, we draw the perpendicular bisector P_{2} of BC. Any point on P_{2} will be equidistant from B and C:

But what about the point of intersection of P_{1} and P_{2}? Since that point lies on P_{1}, it must be equidistant from A and B, and since it also lies on P_{2}, it must be equidistant from B and C as well, which means that it is equidistant from all of A, B and C – this point is therefore the point O – the center of the circle we are seeking to draw! Incidentally, note that O will also lie on the perpendicular bisector of the segment joining A and C, since we have shown that it is equidistant from A and C.

Now, to draw the circle passing through A,B and C, we simply keep the tip of our compass at O, and draw a circle with radius OA (or OB or OC):

We note that this circle is *unique* – no other circle can pass through A, B and C. This is because any point which is equidistant from A,B and C must lie on the perpendicular bisectors of AB and BC (and hence AC as well) – and it is easy to see that O is the only such point.

Also, if A, B and C are collinear, then the perpendicular bisectors of AB and BC would *never intersect* (they would be parallel). Hence, no such point O would exist – there would be no circle passing through all the three points. Let us summarize this discussion in the form of a theorem:

**Theorem:** Given any three non-collinear points A, B and C, there exists a unique circle passing through them.

A number of corollaries can be deduced from this theorem.

**Corollary 1:** If two circles have three points in common, then they must coincide (that is, they must be the same circle).

**Corollary 2:** Two circles cannot intersect in more than two points (this is evident from the previous corollary).

**Corollary 3:** It two circles have a common arc, they must coincide. This is because an arc has infinitely many points, and two distinct circles cannot have more than two points in common.

**Corollary 4:** If A,B and C lie on a circle, and O is a point inside the circle such that OA = OB = OC, then O must be the center of the circle. This is again obvious – there can be only one point which is equidistant from three non-collinear points.