NCERT Solutions for Class 9 Maths Chapter 10 Circles Exercise 10.3

Core Concepts: Circles Intersections and Centers

To solve circles ex 10 3 class 9 ncert, we must understand two major geometric pillars. First, the intersection of circles: when you draw two circles, they can share 0 common points (when separate or nested), exactly 1 common point (when touching externally or internally), or at most 2 common points (when intersecting). They can never intersect at 3 or more points unless they are completely identical (coincident). Second, the perpendicular bisector theorem: the perpendicular drawn from the center of a circle to a chord always bisects that chord. Conversely, the perpendicular bisector of any chord of a circle must pass directly through the circle's center. This mathematical truth allows us to uniquely construct and locate the center of any given circle using simple compass and straightedge techniques.

Step-by-Step Construction: How to Find the Center of a Given Circle

1. Plot Three Points on the Circle — Mark three distinct, non-collinear points anywhere on the circumference of the given circle. Let these points be A, B, and C.
2. Draw Chords AB and BC — Use a straight ruler to draw line segments joining A to B, and B to C. These segments AB and BC act as chords of the circle.
3. Construct Perpendicular Bisectors — Using a compass, draw the perpendicular bisector of chord AB. Similarly, construct the perpendicular bisector of chord BC.
4. Identify the Intersection Point — Extend both perpendicular bisectors until they intersect. The point of intersection, say O, is the unique center of the circle because the perpendicular bisectors of all chords must intersect at the center.

Board Exam Tips & Avoidable Pitfalls

1. Incomplete Construction Lines: When proving or drawing the construction to locate the circle's center, do not erase your arc marks! Examiners look for neat arc intersections to award full step-marks.
2. Confusing Chord Bisectors: Remember that a perpendicular drawn from the center bisects the chord, but an arbitrary line from the center to the chord is not perpendicular unless it bisects it. Keep the condition and its converse clear.
3. Proving Triangles Congruent: In proofs involving intersecting circles, identify the common chord and use RHS or SSS congruency criteria to prove that the centers lie on the perpendicular bisector. Don't skip intermediate steps.

Practice Questions with Solutions

• Q: Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points? A: Step 1: Let us analyze the different ways two circles can be drawn. - Case 1: The circles do not touch or intersect. Number of common points = 0. - Case 2: The circles touch each other externally or internally at a single point. Number of common points = 1. - Case 3: The circles intersect each other. Number of common points = 2. Step 2: Try to draw two circles that intersect at 3 points. Since a unique circle passes through any 3 non-collinear points, two distinct circles cannot have 3 points in common. Final answer: The maximum number of common points between any pair of circles is 2.
• Q: If two circles intersect at two points, prove that their centers lie on the perpendicular bisector of the common chord. A: Step 1: Let two circles with centers O and O' intersect at points A and B. AB is the common chord. Let OO' intersect AB at point M. Step 2: Prove triangles OAO' and OBO' congruent. In ΔOAO' and ΔOBO': - OA = OB (Radii of circle with center O) - O'A = O'B (Radii of circle with center O') - OO' = OO' (Common side) By SSS congruence, ΔOAO' ≅ ΔOBO'. Therefore, ∠AOO' = ∠BOO' (by CPCT), which means ∠AOM = ∠BOM. Step 3: Now compare ΔAOM and ΔBOM: - OA = OB (Radii) - ∠AOM = ∠BOM (Proved above) - OM = OM (Common side) By SAS congruence, ΔAOM ≅ ΔBOM. Hence, AM = BM and ∠AMO = ∠BMO (by CPCT). Step 4: Since ∠AMO + ∠BMO = 180° (linear pair), we have 2∠AMO = 180° => ∠AMO = 90°. Final answer: Since AM = BM and ∠AMO = 90°, OO' is the perpendicular bisector of the common chord AB. Hence, the centers lie on the perpendicular bisector.
• Q: Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centers is 4 cm. Find the length of the common chord. A: Step 1: Let O and O' be the centers of the circles with radii 5 cm and 3 cm respectively. The distance OO' = 4 cm. Let AB be the common chord intersecting OO' at M. Step 2: We know that the line of centers OO' is the perpendicular bisector of AB. Let OM = x. Then O'M = 4 - x. Let AM = y. Step 3: Apply Pythagoras theorem in right ΔAMO and ΔAMO': In ΔAMO: OA² = OM² + AM² => 5² = x² + y² => y² = 25 - x² In ΔAMO': O'A² = O'M² + AM² => 3² = (4 - x)² + y² => y² = 9 - (16 - 8x + x²) Step 4: Equate both equations for y²: 25 - x² = 9 - 16 + 8x - x² 25 = -7 + 8x 32 = 8x => x = 4. Step 5: Substitute x = 4 in the first equation: y² = 25 - 4² = 25 - 16 = 9 => y = 3 cm. Step 6: The length of the common chord AB = 2 AM = 2 y = 2 * 3 = 6 cm. Final answer: The length of the common chord is 6 cm.
• Q: If a line intersects two concentric circles (circles with the same center) with center O at A, B, C, and D, prove that AB = CD. A: Step 1: Draw a perpendicular OM from the center O to the intersecting line AD. Step 2: Since OM is perpendicular to chord AD of the larger outer circle, it bisects AD. Therefore, AM = MD --- (Equation 1) Step 3: Similarly, OM is perpendicular to chord BC of the smaller inner circle, so it bisects BC. Therefore, BM = MC --- (Equation 2) Step 4: Subtract Equation 2 from Equation 1: AM - BM = MD - MC Step 5: Looking at the figure, AM - BM leaves AB, and MD - MC leaves CD. Final answer: Hence, AB = CD. Proved.

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