Question

State the following statement is True or False

A circle can be drawn passing through three non collinear points.

• A. True
• B. False

Answer 1

The correct option is A True

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

Given: Three non collinear points $P,Q,R$

To prove: There is one and only one circle passing through the points $P,Q$ and $R$.

Construction: Join $PQ$ and $QR$.

Draw perpendicular bisectors $AB$ of $PQ$ and $CD$ of $QR$. Let the perpendicular bisectors intersect at the point $O$.

Now join $OP,OQ$ and $OR$.

A circle is obtained passing through the points $P,Q$ and $R$.

Proof: We know that, each and every point on the perpendicular bisector of a line segment is equidistant from its ends points.

Thus, $OP=OQ$ [Since, $O$ lies on the perpendicular bisector of $PQ$]

and $OQ=OR$ [Since, $O$ lies on the perpendicular bisector of $QR$]

So, $OP=OQ=OR$.

Let $OP=OQ=OR=r$

Now draw a circle $C(O,r)$ with $O$ as centre and $r$ as radius

Then circle $C(O,r)$ passes through the points $P,Q,R$

Next, we prove this circle is the only circle passing through the points $P,Q$ and $R$.

If possible, suppose there is a another circle $C(O^{\prime },t)$ which passes through the points $P,Q,R$.

Then, $O\text{'}$ will lie on the perpendicular bisectors $AB$ and $CD$.

But $O$ was the intersection point of the perpendicular bisectors $AB$ and $CD$.

So, $O\text{'}$ must coincide with the point $O$. [Since, two lines can not intersect at more than one point]

As, $O\text{'}P=t$ and $OP=r$ and $O\text{'}$ coincides with $O$, we getv$t=r$

Therefore, $C(O,r)$ and $C(O,t)$ are congruent.

Thus, there is one and only one circle passing through three the given non-collinear points.

hence the statement is correct.