NCERT Class 9 Solutions: Circles (Chapter 10) Exercise 10.3 – Part 1

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Radius, diameter chord, tangent and secant associated with a circle

Radius, Diameter Chord, Tangent and Secant

Radius, diameter chord, tangent and secant associated with a circle

Q-1 Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?

Solution:

  1. No common point.

    Circle c Circle c: Circle through B with center O Circle d Circle d: Circle through D with center O' Point O O = (-1.42, 2.28) Point O O = (-1.42, 2.28) Point O O = (-1.42, 2.28) Point O' O' = (2.22, 2.34) Point O' O' = (2.22, 2.34) Point O' O' = (2.22, 2.34)

    Outer and Inner Circles Which Have No Common Point

    Outer and inner circles with centers O and O' and no common point.

  2. No point common

    Circle c Circle c: Circle through B with center O Circle d Circle d: Circle through C with center O Point O O = (-1.18, 3.58) Point O O = (-1.18, 3.58) Point O O = (-1.18, 3.58) Point O' O' = (-0.86, 3.46) Point O' O' = (-0.86, 3.46) Point O' O' = (-0.86, 3.46)

    Two Circle That Center Point O and O'

    The two circle that center point O and O' and its common point is P and Q

  3. One point P is common

    Circle c Circle c: Circle through B with center O Circle d Circle d: Circle through D with center O' Point O O = (-0.76, 2.88) Point O O = (-0.76, 2.88) Point O O = (-0.76, 2.88) Point O' O' = (2.57, 2.97) Point O' O' = (2.57, 2.97) Point O' O' = (2.57, 2.97) P text1 = "P"

    Circles Which Have a Single Common Point

    Circle with centers O and O' and 1 common point.

  4. One point P is common

    Circle c Circle c: Circle through B with center O Circle d Circle d: Circle through D with center O' Point O O = (0.34, 2.48) Point O O = (0.34, 2.48) Point O O = (0.34, 2.48) Point O' O' = (1.7, 2.52) Point O' O' = (1.7, 2.52) Point O' O' = (1.7, 2.52) P text1 = "P"

    Outer and Inner Circles Which Have a Single Common Point

    Outer and inner circles with centers O and O' and 1 common point.

  5. Two point P and Q are common

    Circle c Circle c: Circle through B with center O Circle d Circle d: Circle through D with center O' Point O O = (-0.54, 3.22) Point O O = (-0.54, 3.22) Point O O = (-0.54, 3.22) Point O' O' = (3.22, 3.32) Point O' O' = (3.22, 3.32) Point O' O' = (3.22, 3.32) P text1 = "P" Q text2 = "Q"

    Circles Which Have a 2 Common Points

    Circle with centers O and O' and 2 common point P and Q.

  6. Infinite number of common points between two congruent and overlapping circles

    Circle c Circle c: Circle through B with center O Circle d Circle d: Circle through C with center O Point O O = (-1.52, 4) Point O O = (-1.52, 4) Point O O = (-1.52, 4) Point O' O' = (-1.54, 4) Point O' O' = (-1.54, 4) Point O' O' = (-1.54, 4)

    Two Congruent and Overlapping Circles With Infinite Common Points

    Two congruent and overlapping circles O and O' and infinite common point P and Q.

    Therefore there can be infinite such points

Q-2 Suppose you are given a circle. Give a construction to find its centre.

Solution:

Circle c Circle c: Circle through B with center A Arc d Arc d: CircumcircularArc[I, J, K] Arc e Arc e: CircumcircularArc[L, M, N] Arc k Arc k: CircumcircularArc[O, P, Q] Arc p Arc p: CircumcircularArc[R, S, T] Arc q Arc q: CircumcircularArc[U, V, W] Arc r Arc r: CircumcircularArc[Z, A_1, B_1] Arc s Arc s: CircumcircularArc[C_1, D_1, E_1] Arc t Arc t: CircumcircularArc[F_1, G_1, H_1] Segment f Segment f: Segment [A, C] Segment g Segment g: Segment [A, D] Segment h Segment h: Segment [E, F] Segment i Segment i: Segment [G, H] Point A A = (0.46, 2.66) Point A A = (0.46, 2.66) Point A A = (0.46, 2.66) Point E Point E: Point on c Point E Point E: Point on c Point E Point E: Point on c Point F Point F: Point on c Point F Point F: Point on c Point F Point F: Point on c Point G Point G: Point on c Point G Point G: Point on c Point G Point G: Point on c Point H Point H: Point on c Point H Point H: Point on c Point H Point H: Point on c

Circle With Center at a

Circle at center A and EF and GH are two chords.

Key idea: The key idea for solving this problems is to realize that the perpendicular bisector of chords of the circle pass through the center. So if we draw two perpendicular bisectors of two distinct chords they will intersect at the center of the circle. (The line joining the center and perpendicular to the chord, bisects the chord)

Following steps can be used to complete the construction:

  • Step 1 : Drawn the circle

  • Step 2 : EF and GH are two chords

  • Step 3: Draw the perpendicular bisectors of the chords EF and GH.

The point of intersection of two perpendicular bisector is the center of the circle.

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