NCERT Solutions Class 9 Maths Chapter 10: Circles Exercise 10.6

Theoretical Foundations of Advanced Circle Proofs

To successfully solve class 9 maths circles ex 10 6 problems, you must be comfortable synthesizing multiple core theorems simultaneously. The exercise heavily depends on the relationships between overlapping circles, congruent triangles, and cyclic quadrilaterals. Remember, a quadrilateral is cyclic if and only if its opposite angles sum to 180 degrees. Additionally, the angle subtended by an arc at the center is always double the angle subtended by it at any point on the remaining part of the circle. When tackling complex multi-circle problems, look for shared chords or perpendicular bisectors. Utilizing the property that the perpendicular from the center of a circle to a chord bisects the chord is vital for set-up equations.

Systematic Proof Strategy for Exercise 10.6

1. Construct an Accurate Diagram — Always begin by drawing a neat diagram. Clearly mark given lengths, congruent parts, centers of the circles, and angles. Use a compass to ensure correct relative sizing of overlapping circles.
2. Identify Auxiliary Elements — Many advanced proofs require you to draw extra lines. Try joining the centers of intersecting circles, connecting centers to chord endpoints, or dropping perpendiculars from the center to a chord.
3. Apply Triangle Congruence Criteria — Look for pairs of triangles that can be proven congruent using SSS, SAS, or RHS. Congruence is the primary tool to prove that corresponding parts of circles (like chord segments) are equal (CPCT).
4. Utilize Angle Tracking & Cyclic Properties — Track angle relationships across segments. If you need to prove a shape is cyclic, search for angles in the same segment that are equal, or show that opposite angles are supplementary.

Exam Tips: Avoid Common Pitfalls in Circle Proofs

1. Don't Assume Concollicity: Never assume four points lie on a circle (are concyclic) unless you prove that the opposite angles of the quadrilateral they form add up to 180 degrees, or that angles subtended by a line segment at two other points on the same side are equal.
2. Quote the Exact Theorems: CBSE examiners look for precise theorem names in brackets. Instead of writing 'by theorem', write 'Perpendicular from the center to a chord bisects the chord'.
3. Coordinate Axes Fallacy: Do not use coordinate geometry methods to solve these purely Euclidean geometry proofs unless specifically instructed. Stick to geometric theorems.

Practice Questions with Solutions

• 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 the two circles have centers O and O' with radii 5 cm and 3 cm respectively. Let the distance OO' = 4 cm. Step 2: Let the common chord be AB, intersecting OO' at point M. Since the line joining centers of two intersecting circles is the perpendicular bisector of their common chord, we have AM = MB and OO' is perpendicular to AB. Step 3: Let OM = x. Since OO' = 4 cm, O'M = 4 - x. From right triangle AMO: $AM^2 = OA^2 - OM^2 = 5^2 - x^2 = 25 - x^2$. From right triangle AMO': $AM^2 = O'A^2 - O'M^2 = 3^2 - (4-x)^2 = 9 - (16 - 8x + x^2) = -7 + 8x - x^2$. Step 4: Equate both expressions for $AM^2$: $25 - x^2 = -7 + 8x - x^2$ $32 = 8x \implies x = 4$ cm. Step 5: Substitute $x=4$ back to find AM: $AM^2 = 25 - 4^2 = 9 \implies AM = 3$ cm. Since AB is the common chord, $AB = 2 \times AM = 2 \times 3 = 6$ cm. Final answer: The length of the common chord is 6 cm.
• Q: If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord. A: Step 1: Let AB and CD be two equal chords of a circle with center O, intersecting at point P. We need to prove AP = CP and BP = DP. Step 2: Draw perpendiculars OL from center O to AB, and OM to CD. Join OP. Step 3: Compare right triangles OLP and OMP: - $OL = OM$ (Equal chords of a circle are equidistant from the center) - $\angle OLP = \angle OMP = 90^\circ$ (By construction) - $OP = OP$ (Common hypotenuse) Therefore, $\triangle OLP \cong \triangle OMP$ by RHS congruence criterion. Step 4: By CPCT, we get $LP = MP$. Also, since perpendicular from center bisects the chord, $AL = LB = \frac{1}{2}AB$ and $CM = MD = \frac{1}{2}CD$. Since $AB = CD$, we have $AL = CM$ and $LB = MD$. Step 5: To prove AP = CP, add LP to AL and MP to CM: $AL + LP = CM + MP \implies AP = CP$. To prove BP = DP, subtract LP from LB and MP from MD: $LB - LP = MD - MP \implies BP = DP$. Final answer: Hence proved, the corresponding segments of the intersecting equal chords are equal.
• Q: Prove that the non-parallel sides of an isosceles trapezium are equal, and show that any isosceles trapezium is cyclic. A: Step 1: Let ABCD be an isosceles trapezium where AB is parallel to CD, and non-parallel sides AD = BC. Step 2: Draw perpendiculars AM and BN from A and B respectively to the line CD. Step 3: In right triangles AMD and BNC: - $AM = BN$ (Distance between parallel lines AB and CD is constant) - $\angle AMD = \angle BNC = 90^\circ$ - $AD = BC$ (Given non-parallel sides are equal) By RHS congruence, $\triangle AMD \cong \triangle BNC$. Step 4: By CPCT, we have $\angle D = \angle C$. Also, since AB is parallel to CD, the consecutive interior angles must sum to 180 degrees: $\angle A + \angle D = 180^\circ$. Step 5: Substitute $\angle D = \angle C$ into the angle equation: $\angle A + \angle C = 180^\circ$. Since the sum of a pair of opposite angles of quadrilateral ABCD is 180 degrees, the quadrilateral is cyclic. Final answer: Hence proved, any isosceles trapezium is a cyclic quadrilateral.
• Q: If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the segment, prove that the four points lie on a circle. A: Step 1: Let AB be a line segment. Let C and D be two points on the same side of AB such that $\angle ACB = \angle ADB$. Step 2: We must prove that points A, B, C, and D are concyclic. Let us assume they do not lie on the same circle. Step 3: Draw a circle passing through any three non-collinear points A, B, and C. Suppose this circle does not pass through D. Then it must intersect AD (or AD produced) at some point D'. Step 4: Join D'B. Since points A, B, C, and D' lie on the circle, we have: $\angle ACB = \angle AD'B$ (Angles subtended by the same arc AB are equal). But we are given that $\angle ACB = \angle ADB$. Therefore, $\angle AD'B = \angle ADB$. Step 5: In triangle BDD', an exterior angle ($\angle AD'B$) cannot be equal to the interior opposite angle ($\angle ADB$) unless the points D and D' coincide. Thus, our assumption was wrong. Point D must lie on the circle. Final answer: Hence, the four points A, B, C, and D are concyclic.

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