Question

refer this second attachment​

Answer 1

Step-by-step explanation:

☆Hint:☆

It is given that there are three non-collinear points in a circle that is A, B and C.

We have to prove that there is one and only one circle passing through the points A, B and C.

Construction: As we have marked three non-collinear points in circle. Now, let’s join AB and BC. Then, Draw the perpendicular bisectors of RS and PQ of the chords AB and BC respectively as shown in figure.

Let us assume PQ and RS intersect in O. Here, Join OA, OB and OC.

Let’s proof: The point O lies on the perpendicular bisector of AB.

Therefore, OA = OB

And also, O lies on the perpendicular bisector of BC.

Therefore, OB = OC

Thus, we can say from above OA = OB = OC = r (where r stands for radius of the circle).

By taking O as the centre, construct a circle of radius r C(0,r)

which passes through the points A, B and C.

Let us suppose there is another circle with the centre O′

and radius r, passing through the A, B and C. Then, O′

will also lie on the perpendicular bisector of PQ and RS.

As we know that, the two lines cannot intersect at more than one point. So, O′

must coincide with O which means O′

and O lies at the same point.

Hence, there is one and only one circle passing through three non collinear points that are A, B and C.

Hence Proved.

Note:

To solve these types of questions, you must construct a figure to solve the question. Here, you can also suppose or can take assumptions for better understanding in a practical way.

Answer 2

Explanation:-

$\star\;{\underline{\boxed{\pmb{\green{\frak{\;Given\; : -}}}}}}$

• Three non collinear points A,Band C

$\star\;{\underline{\boxed{\pmb{\green{\frak{\;To\:Find\; : -}}}}}}$

• One and only circle passes through A,B and C.

$\star\;{\underline{\boxed{\pmb{\pink{\frak{\; Solution\; : -}}}}}}$

Proof : Since O lies on perpendicular bisector of AB

$\:\:\:\:\:\:\:\:\:\huge\rightarrow$OA = OB

Since , O lies on perpendicular bisector of BC

$\:\:\:\:\:\:\:\:\:\huge\rightarrow$OB = OC

$\:\:\:\:\:\:\:\:\:\huge\rightarrow$OA = OB = OC

O is equivalent from A,B and C .

If a circle is drawn with centre O and radius OA , the circle will pas s through B and C and Since perpendicular bisectors of AB and BC cut each other at point O only.

O is the only pint equivalent from A,Band C.

Therefore, one and only one circle be drawn through three non collinear points .

$\color{Red}\boxed{Additional\: Information}$

1. Angle in alternate segments are equal.

2. Angle in same segments are equal.

3. Angle subtended at the centre of a circle by an arc is double the angle subtended on the circumference by the same arc .

4. Sum of the opposite pair of angles of cyclic quadrilateral is supplementary.

5. Exterior angle of a cyclic quadrilateral is equals to interior opposite angle