The collection of all points in plane,which are at fixed distance from a fixed points in the plane, is called circle. And the the fixed point is called center of the circle and the fixed distance is called the radius of the circle. Note that the line segment joint the center and any points on the circle is also called the radius of the circle.
The circle have 3 parts. One is the outside the circle is called exterior of the circle. Second in side the circle which is called interior of the circle. Third is the circle itself. The circle and its interior will make up the circular region. if we take 2 points on the circle, P and Q, then the line segment PQ is called chord of the circle. The chord which is passing through the center of the circle is called diameter of the circle. The diameter is the biggest chord in the circle. And all the diameter in the circle is same. Also the diameter is the 2 times of the radius of the circle. We can also see that we can only draw maximum 4 numbers of diameter in a circle.
A piece of the circle between 2 points on the circle is called arc of the circle. The arc of the circle can be 2 types. 1 is minor arc and major arc, depending upon there length. If a circle have 2 same arc in there length , then it will called semi circle. The length of the complete circle is called the circumference of the circle. The region between a chord and either of its arc is called segment of the circle. Segment can be either major or minor segment, depending upon there arc length as shown in the fig. the region between an arc and two radii joining the center to the two end points of the arc is called sector of the circle. It can also be Major sector and minor sector depending upon the arc length as shown in the fig.
See p-171 for the exercise 10.1
Angle subtended by a chord at a point
Consider the line segment PQ as the chord of the circle as shown in the figure and the point R on the circle. Then the angle PRQ is called the angle subtended by the chord PQ at the point R. The angle POQ is the angle subtended by the chord PQ at the center of the circle. We can also notice the major and minor arc PQ in fig.
Now we can examine the relation between the length of the chord and angle subtended by them at the center.
In the above fig. line segment AB, CD, EF are the equal chords of the circle. Now consider the triangle ABO and CDO . The sides AB and CD are same (given), side AO and CO are also same as both are radius of the same circle and the side BO and DO are also same as they are the radius of the same circle. So we can understand that triangle ABO and triangle CDO are congruent triangle (sss congruent rule). There for the angle AOB and COD are same (CPCT). So we can state this observation as a theorem.
Theorem 10.1. Equal chord of a circle subtend equal angle at the center.
Theorem 10.2 is just converse of the 10.1
Theorem 10.2 . If the angles subtended by the chords of a circleat the center are equal, then the chords are equal.
This one can also prove with the same figure. Consider the triangle ABO and CDO and take the angles AOB and DOC are same. Now we can say that the side OA and side DO are the same because they are the radius of the same circle. Same way we can say that side OC and OB are same as it also the radius of the same circle. Now we can say both triangle are congruent by S(AS). There for the side AB and CD are same (CPCT). See p-173 for the exercise 10.2
perpendicular form the center to a chord
As shown in the figure consider the chord AB and the perpendicular OM. Now we can join the OA and OB, so that we will get two right angle triangle. We can see that side OA and OB are same, as it is the radius of the same circle. And the side OM is common for both triangle. Now we can say that the both triangle formed is a congruent triangles (RHS). Ther for we can say that AM =BM. And it can be conclude as a theorem
Theorem 10.3 The perpendicular from the center of a circle to a chord bisects the chord.
The covers of this theorem is
Theorem 10.4 The line drawn through the center of a circle to bisect a chord is perpendicular to the chord.
we can prove this by the following figure.
Here AM=BM (Given), OA = OB 9 (radius), OM=OM (cmone side). So all three sides of this triangle AOM and MOB are same, and these triangle are congruent by (sss). There for angle AMO = BMO=90° . Since the linear angle AB IS 180. So OM⊥ AB.