# Circle Passing through 3 Points: Circle Passing through a Point (For CBSE, ICSE, IAS, NET, NRA 2022)

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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, letΥs 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.