# Circle Passing through 3 Points: Circle Passing through a Point

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To draw a straight line the minimum number of points required is two. Given any two points, a straight line can be drawn. How many minimum points are sufficient to draw a unique circle? Well, let’s try to find out.

## Circle Passing through a Point

Let us consider a point and try to draw circle passing through that point.

It can be seen that through a single point P infinite circles passing through it can be drawn.

## Circle Passing through Two Points

Now, let us take two points P and Q and see what happens?

Again, we see that an infinite number of circles passing through points P and Q can be drawn.

## Circle Passing through Three Points (Collinear or Non-Collinear)

Let us now take 3 points. For a circle passing through 3 points, two cases can arise.

• Three points can be collinear

• Three points can be non-collinear

Let us study both cases individually.

### Case 1: A Circle Passing through 3 Points: Points Are Collinear

Consider three points P, Q and R which are collinear.

Here, seen that if three points are collinear any one of the points either lie outside the circle or inside it. Therefore, a circle passing through 3 points, where the points are collinear is not possible.

### Case 2: A Circle Passing through 3 Points: Points Are Non-Collinear

To draw a circle through three non-collinear points join the points as shown:

Draw perpendicular bisectors of PQ and RQ. Let the bisectors AB and CD meet at O.

With O is the centre and radius OP or OQ or OR draw a circle. We get a circle passing through 3 point P, Q, and R.

It is observed that only a unique circle will pass through all the three points. It can be stated as a theorem and the proof is explained as follows.

Given: Three non-collinear points A, B and C

To prove: One and only one circle can be drawn through A, B, and C

Construction: Join AB and BC. Draw perpendicular bisectors of AB and BC. Let these perpendicular bisectors meet at a point O.

### Proof

 Sr. No. Statement Reason 1. Every points of perpendicular bisector is equidistant from the extremities of line. 2. Every points of perpendicular bisector is equidistant from the extremities of line. 3. From statement (1) and (2). 4. O is equidistant from A, B and C Of a circle is drawn with O as centre and OA as radius, then it will also pass through B and C. O is the only point equidistant from A, B and C as the perpendicular bisector of AB and BC intersect at point O only.

From above it follows that a unique circle passing through 3 points can be drawn given that the points are non-collinear.