Question

Prove that there is one and only one circle passing through three given non-collinear points.

Answer 1

Given : Three non - collinear points, $P,Q$ and $R.$
To prove : There is one and only one circle passing through the points $P,Q$, and $R.$
Construction: Draw line segments $PQ$ and $QR.$ Draw perpendicular bisectors MN and $ST$ of $PQ$ and $RQ$ respectively. Since $P,Q,R$ are not collinear, $MN$ is not parallel to $ST$ and will intersect, at the point $O.$ Join $OP,OQ$ and $OR$ (Fig).

Proof :

As $O$ lies on $MN$, the perpendicular bisector of $PQ$,

In $△OXP$ and $△OXQ,$ we have

$OX=OX$ [Common]

$\angle OXP=\angle OXQ$ [Right angles]

$XP=XQ$ [$MN$ is the perpendicular bisector]

$\therefore △OXP\cong △OXQ$ [By SAS congruence Rule]

$OP=OQ$ [CPCT]

Similary, $△OYQ\cong △OYR$

and $OQ=OR$ [CPCT]

$\therefore OP=OQ=OR=r\left(suppose\right)$

Taking $O$ as the centre and $r$ as the radius, draw a circle $C(O,r)$ which will pass through $P,Q$, and $R.$

If possible, suppose there is another circle $C(O^{\prime },r^{\prime })$ passing through $P,Q$ and $R.$ Then $O^{\prime }$ will lie on the perpendicular bisector $MN$ of $PQ$ and $ST$ of $QR.$

Since two lines cannot intersect at more than one point, $O^{\prime }$ must coincide with $O.$ Since $OP=r,O^{\prime }P=r^{\prime }.$
We have, $r=r^{\prime }$
Hence, $C(O^{\prime },r^{\prime })=C(O,r)$

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