Question

There is one and only one circle passing through three given non collinear points solution

Answer 1

Answer:

Step-by-step explanation:

Let us take three points A, B and C, which are not on the same line, or in other words, they are not collinear [as in figure]. Draw perpendicular bisectors of AB and BC say, PQ and RS respectively. Let these perpendicular bistros intersect at one point O. (Note that PQ and RS will intersect because they are not parallel) [as in figure].

∴ O lies on the perpendicular bisector PQ of AB.  

∴ OA = OB  

[∵ Every point on the perpendicular bisector of a line segment is equidistant from its end points]  

Similarly, ∴ O lies on the perpendicular bisector RS of BC. ∴ OB = OC  

[∵ Every point on the perpendicular bisector of a line segment is equidistant from its end points]  

So, OA = OB = OC  

i.e., the points A, B and C are at equal distances from the point O.  

So, if we draw a circle with centre O and radius OA it will also pass through B and C. This shows that there is a circle passing through the three points A, B and C. We know that two lines (perpendicular bisectors) can intersect at only one point, so we can draw only one circle with radius OA. In other words, there is a unique circle passing through A, B and C.  

Answer 2

Statement : There is one and only one circle passing through three given noncollinear points.

Given : AB and CD are two equal chords of the circle.

OM and ON are perpendiculars from the centre at the chords AB and CD.

To prove : OM = ON.
Construction : Join OA and OC.

Proof :

In ΔAOM and ΔCON,

OA = OC . (radii of the same circle)
MA = CN . (since OM and ON are perpendicular to the chords and it bisects the chord and AM = MB, CN = ND)

∠OMA = ∠ONC = 90°
ΔAOM ≅ ΔCON (R. H. S)
OM = ON (c. p. c. t.)

Equal chords of a circle are equidistant from the centre.