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Circles

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Circles

1. 1. CHAPTER 10
2. 2. SOME DEFINITIONS • CIRCLES- A circle is a collection of those points in a plane that are at a given constant distance from a given fixed point in the plane. r O X
3. 3. The collection of all points lying inside and on the circle C(O , r) is called a circular disc with centre O and radius r. Circles having the same centre but with different radii are said to be concentric circles. . O
4. 4. A continuous piece of a circle is called an arc of the circle. r O. P6 P2 P4 P5 P1 P3 Consider a circle C(O , r). Let P1, P2, P3, P4, P5, P6 be points on the circle. Then, the pieces P1,P2,P3,P4,P5,P6 etc. are all arcs of the circle C(O, r)
5. 5. Arc Minor Arc – A minor arc of a circle is the collection of those points that lie on and also inside a central angle. Major Arc - A major arc of a circle is the collection of points of the circle that lie on or outside a central angle
6. 6. CHORD :- A line segment joining any two points on a circle is called a chord of the circle. A chord passing through the centre of a circle is known as its diameter. P Q r rO P Q QP PQ d
7. 7. THEOREM’S Given- Arc PQ of a circle C(O , r) and arc RS of another circle C(O’ , r) congruent to C(O , r) such that PQ ≅ RS. To Prove- PQ=RS Construction- Draw line segments OP, OQ, O’R and O’S If two arcs of a circle are congruent , then corresponding chords are equal. O P Q O’ R S
8. 8. Proof :- In triangles OPQ and O’RS , we have OP=OQ=O’R=O’s=r [Equal radii of two circles] ∠ POQ = ∠ RO’S [∵ PQ ≅ RS m(PQ)= m(RS) ∠POQ= ∠RO’S] So, by SAS criterion of congruence , we have ▲POQ ≅ ▲RO’S PQ=RS If PQ , RS are major arcs , then QP and SR are minor arcs. So, PQ ≅ RS , PQ=RS QP=SR , QP=SR Hence, PQ ≅ RS  PQ=RS
9. 9. Given- A chord PQ of a circle C(O, r) and perpendicular OL to the chord PQ. To Prove- LP=LQ Construction- Join OP and OQ. O P Q L Proof- In triangles PLO and QLO , we have OP=OQ=r [Radii of the same circle] OL= OL [Common] And, ∠OLP= ∠OLQ [Each equal to 90°]
10. 10. THEOREM 3 :- Given - A chord PQ of a circle C(O, r) with mid-point M. To prove – OM |PQ Construction - Join OP and OQ Proof – In triangles OPM and OQM, we have OP=OQ [Radii of the same circle] PM=MQ [M is mid-point of PQ] OM=OM So, SSS-criterion of congruence , we have ▲OPM ≅ ▲OQM  ∠OMP= ∠OMQ But, ∠OMP+ ∠OMQ=180º  ∠OMP+ ∠OMP=180º  2∠OMP=180º  ∠OMP=90º  ∠OMP=∠OMQ=90º [∵ ∠OMP= ∠OMQ] Hence, OM |PQ O P Q M
11. 11. Given- Three non-collinear points P,Q and R. To Prove- There is one and only one circle passing through P, Q and R. Construction- Join PQ and QR. Draw perpendicular bisector AL and BM of PQ and RQ respectively. Since P,Q,R are not collinear. Therefore, the perpendicular bisectors AL and BM are not parallel. Let AL and BM intersect at O. Join OP,OQ and OR. R A B M QP L O
12. 12. Proof – Since O lies on the perpendicular bisector of PQ. Therefore, OP=OQ , Again, O lies on the perpendicular bisector of QR. Therefore, OQ=OR Thus OP=OQ=OR=r . Taking O as the centre draw a circle of radius s. Clearly, C(O, s) passes through P, Q and R. This proves that there is a circle passing through the point P, Q and R. We shall now prove that this is the only circle passing through P, Q and R. Let there be another circle with centre O’ and radius r, passing through the points P,Q and R. Then, O’ will lie on the perpendicular bisectors AL of PQ and BM of QR. Since two lines cannot intersect at more than one point , so O’ must coincide with O. Since OP=r, O’P=s and O and O’ coincide, we must have r=s  C(O, r) ≅ C(O, s)  Hence, there is one and only one circle passing through three non- collinear points P, Q and R.
13. 13. Ans 1: The below given steps will be followed to find the centre of the given circle. Step1. Take the given circle. Step2. Take any two different chords AB and CD of this circle and draw perpendicular bisectors of these chords. Step3. Let these perpendicular bisectors meet at point O. Hence, O is the centre of the given circle.
14. 14. Ans 2: Consider two circles centered at point O and O’, intersecting each other at point A and B respectively. Join AB. AB is the chord of the circle centered at O. Therefore, perpendicularbisector of AB will pass through O. Again, AB is also the chord of the circle centered at O’. Therefore, perpendicular bisector of AB will also pass through O’. Clearly, the centers of these circles lie on the perpendicular bisector of the common chord.
15. 15. Q 3: Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points? Ans 3: Consider the following pair of circles. The above circles do not intersect each other at any point. Therefore, they do not have any point in common.
16. 16. The below circles touch each other at 1 point X only. Therefore, the circles have 1 point in common.
17. 17. These circles intersect each other at two points G and H. Therefore, the circles have two points in common. It can be observed that there can be a maximum of 2 points in common. Consider the situation in which two congruent circles are superimposed on each other. This situation can be referred to as if we are drawing the circle two times.