Chapter 10 Circles Exercise 10.3
Question 1: Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?Answer:
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjy2u5op-0uL8QmirQfs_RzbMK5_TiYqi3b6FYDWnYW3tTGji98AcJZscFYrGInfjomCD1s9lXBhSZosy6j1qGkZqG05pG7w9ITkSnKcSlG3uaXYvqemgebO41w1NrHny7XcI__ZYLdtauy/s16000/10.3.png)
Therefore, the maximum number of common points is 2.
Question 2: Suppose you are given a circle. Give a construction to find its centre.
Answer:
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhsgy1ueABNUBkSn6N_nMqnLC7e6YHkjnJpWUN3KKHaZS2_oidk4C1LPSSZrROFec9pChb79yVGujpU6oMrvgbs5UZoykZAazCI0PQVN606GbF3BZef-5m8PSaY-x71cri8AbKPr8DitJVn/s16000/10.3+%25281%2529.png)
1. Trace a circle through any circular object.
2. Mark three 3 non-collinear points on the circle - A, B and C and join them.
3. Draw the perpendicular bisectors of AB and AC.
4. We observe that the perpendicular bisectors intersect at a point (O).
5. We know that every point on the perpendicular bisector of a line segment is equidistant from the endpoints.
6. Therefore the point of intersection (O) of the perpendicular bisectors indicates the centre of the circle.
Question 3: If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.
Answer:
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhZIeegkH-TXyKz6kZ8VN2X_A9m8aSOraJsRAjwaPiRkk2Y8hudAtfLrgkXo9AO1YWfB7R00nvYIdQ7ZSsbSnMispXzHZOBgSCt9bDjRByVe79MyUV3yJuQKGpIgzMECRdd2Nfmn5_VVwRk/s16000/10.3.jpg)
Circles with centres C and D (intersecting circles). AB is the common chord.
Required to prove:
CD is the perpendicular bisector of AB.
Proof:
Join AC, BC, BD, AD. Let AB and CD intersect at O.
Considering ∆ACD and ∆BCD
CD = CD (common)
AC = BC (radius of the same circle)
AD = BD (radius of the same circle)
Therefore by SSS congruency, ∆ACD ≅ ∆BCD.
By CPCT ∠ACD = ∠BCD.
Again considering ∆ACO and ∆BCO
CO = CO (common)
AC = BC (radius of the same circle)
∠ACD = ∠BCD (radius of the same circle)
Therefore by SAS congruency, ∆ACO ≅ ∆BCO.
By CPCT, AO = BO and ∠AOC = ∠BOC.
But ∠AOC = ∠BOC form a linear pair.
∠AOC + ∠BOC = 180°
2∠AOC = 180°
∠AOC = 90°
Hence CD bisects AB perpendicularly.
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