You have learnt in Chapter 6, that two points are sufficient to determine a line. That is, there is one and only one line passing through two points. A natural question arises. How many points are sufficient to determine a circle?
Take a point P. How many circles can be drawn through this point? You see that there may be as many circles as you like passing through this point [see Fig. 10.17(i)].
Now take two points P and Q. You again see that there may be an infinite number of circles passing through P and Q [see Fig.10.17(ii)]. What will happen when you take three points A, B and C? Can you draw a circle passing through three collinear points?
No. If the points lie on a line, then the third point will lie inside or outside the circle passing through two points (see Fig 10.18).
So, let us take three points A, B and C, which are not on the same line, or in other words, they are not collinear [see Fig. 10.19(i)]. Draw perpendicular bisectors of AB and BC say, PQ and RS respectively.
Let these perpendicular bisectors intersect at one point O. (Note that PQ and RS will intersect because they are not parallel) [see Fig. 10.19(ii)].
Now O lies on the perpendicular bisector PQ of AB, you have OA = OB, as every point on the perpendicular bisector of a line segment is equidistant from its end points, proved in Chapter 7.
Similarly, as O lies on the perpendicular bisector RS of BC, you get
`OB = OC`
So OA = OB = OC, which means that the points A, B and C are at equal distances from the point O. So if you 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. You know that two lines (perpendicular bisectors) can intersect at only one point, so you can draw only one circle with radius OA.
In other words, there is a unique circle passing through A, B and C. You have now proved the following theorem:
`color{orange}("Theorem 10.5 :")"There is one and only one circle passing through three given non-collinear points."`
`color{orange}("Remark :")` If ABC is a triangle, then by Theorem 10.5, there is a unique circle passing through the three vertices A, B and C of the triangle. This circle is called the circumcircle of the `Delta` ABC.
Its centre and radius are called respectively the circumcentre and the circumradius of the triangle.
You have learnt in Chapter 6, that two points are sufficient to determine a line. That is, there is one and only one line passing through two points. A natural question arises. How many points are sufficient to determine a circle?
Take a point P. How many circles can be drawn through this point? You see that there may be as many circles as you like passing through this point [see Fig. 10.17(i)].
Now take two points P and Q. You again see that there may be an infinite number of circles passing through P and Q [see Fig.10.17(ii)]. What will happen when you take three points A, B and C? Can you draw a circle passing through three collinear points?
No. If the points lie on a line, then the third point will lie inside or outside the circle passing through two points (see Fig 10.18).
So, let us take three points A, B and C, which are not on the same line, or in other words, they are not collinear [see Fig. 10.19(i)]. Draw perpendicular bisectors of AB and BC say, PQ and RS respectively.
Let these perpendicular bisectors intersect at one point O. (Note that PQ and RS will intersect because they are not parallel) [see Fig. 10.19(ii)].
Now O lies on the perpendicular bisector PQ of AB, you have OA = OB, as every point on the perpendicular bisector of a line segment is equidistant from its end points, proved in Chapter 7.
Similarly, as O lies on the perpendicular bisector RS of BC, you get
`OB = OC`
So OA = OB = OC, which means that the points A, B and C are at equal distances from the point O. So if you 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. You know that two lines (perpendicular bisectors) can intersect at only one point, so you can draw only one circle with radius OA.
In other words, there is a unique circle passing through A, B and C. You have now proved the following theorem:
`color{orange}("Theorem 10.5 :")"There is one and only one circle passing through three given non-collinear points."`
`color{orange}("Remark :")` If ABC is a triangle, then by Theorem 10.5, there is a unique circle passing through the three vertices A, B and C of the triangle. This circle is called the circumcircle of the `Delta` ABC.
Its centre and radius are called respectively the circumcentre and the circumradius of the triangle.