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# Theorem – There is one and only one circle passing through three given non-collinear points | Class 9 Maths

• Last Updated : 06 May, 2021

### Theorem Statement: There is one and only one circle passing through three given non-collinear points.

Required Diagram:

Given: Three non-collinear points P, Q, and R.

To Prove: There is one and only one circle passing through P, Q, and R.

Construction: Join PQ and QR. Draw perpendicular bisectors AL and BM of PQ and RQ respectively. Since P, Q, R is not collinear. Therefore, the perpendicular bisectors AL and BM are not parallel. Let AL and BM intersect at O. Join OP, OQ, and OR.

### Proof

Since O lies on the perpendicular bisector of PQ. Therefore,

OP = OQ

Again, O lies on the perpendicular bisector of QR. Therefore,

OQ = OR

Thus, OP = OQ = OR = r (say).

Taking O as the centre draw a circle of radius s. Clearly, C (0, s) passes through P, Q, and R. This proves that there is a circle passing through the points P, Q, and R. We shall now prove that this is the only circle passing through P. Q and R. If possible, let there be another circle with centre Oâ€™ and radius r, passing through the points P, Q and R. Then, Oâ€™ will lie on the perpendicular bisectors AL of PQ and BM of QR. Since two lines cannot intersect at more than one point, so Oâ€™ must coincide with O. Since OP = r, Oâ€™P = s and O and Oâ€™ coincide, which means that,

r = s

Therefore,

C(O, r) â‰… (Oâ€™, s)

Hence, there is one and only one circle passing through three non-collinear points P, Q and R.

### Examples

Example 1: Write down the step-by-step construction procedures to find out the centre of the circle?

Solution:

Let the circle be C1. We need to find its centre.

Step 1: Take points P, Q, R on the circle

Step 2: Join PR and RQ.

We know that the perpendicular bisector of a chord passes through the centre. So, we construct perpendicular bisectors of PR and RQ.

Step 3: Take a compass. With point P as the pointy end and R as the pencil end of the compass, mark an arc above and below PR. Do the same with R as pointy end P as pencil end of the compass.

Step 4: Join points intersected by the arcs. The line formed is the perpendicular bisector of PR.

Step 5: Take a compass, with point R as the pointy end and Q as pencil end of the compass mark an arc above and below RQ. Do the same with Q as the pointy end and R as the pencil end of the compass.

Step 6: Join the points intersected by the arcs. The line formed is the perpendicular bisector of RQ.

Step 7: The point where two perpendicular bisectors intersect is the centre of the circle. Mark it as point O. Thus, O is the centre of the given circle.

Example 2: Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?

Solution:

(i) No point common:

(ii) One point common:

(iii) Two points common:

As we can analyze from above, two circles can cut each other maximum at two points.

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