Circles: High-Weightage 4-Mark and 5-Mark Board Riders Class 10

In CBSE Class 10 Mathematics, Chapter 10 (Circles) has the shortest NCERT theory (only two theorems: Theorem 10.1 and Theorem 10.2), yet it generates some of the most rigorous and elegant 4-mark and 5-mark geometric riders on the board exam.

These riders test whether you can synthesize Theorem 10.2 (equal tangent lengths from an external point) with cyclic quadrilaterals, angle subtensions, and triangle congruence. Every board exam paper features at least one of five canonical circle riders.

In this guide, we provide complete, examiner-approved formal proofs for the most famous circle riders.

What You Will Learn

• Formal proof: Circumscribing Quadrilateral Theorem ($AB + CD = AD + BC$)
• Formal proof: A parallelogram circumscribing a circle is a rhombus
• Formal proof: Opposite sides of a circumscribed quadrilateral subtend supplementary angles at the center
• Formal proof: Angle between two tangents is twice the chord angle: $\angle PTQ = 2\angle OPQ$
• Formal proof: The parallel tangents rider ($\angle AOB = 90^\circ$)
• Marking rubrics and presentation guidelines

1. Rider 1: Circumscribing Quadrilateral ($AB + CD = AD + BC$)

Problem: A quadrilateral $ABCD$ is drawn to circumscribe a circle. Prove that: $\mathbf{AB + CD = AD + BC}$

                                  A ------ P ------ B
                                 /                                                  S                    Q
                               /                                                    D -------- R ----------- C

Formal Proof:

1. Given: A quadrilateral $ABCD$ circumscribing a circle with center $O$, touching sides $AB, BC, CD,$ and $DA$ at points $P, Q, R,$ and $S$ respectively.
2. To Prove: $AB + CD = AD + BC$.
3. Proof: By Theorem 10.2, the lengths of tangents drawn from an external point to a circle are equal:
   • From vertex $A$: $AP = AS \quad \text{--- (1)}$
   • From vertex $B$: $BP = BQ \quad \text{--- (2)}$
   • From vertex $C$: $CR = CQ \quad \text{--- (3)}$
   • From vertex $D$: $DR = DS \quad \text{--- (4)}$
4. Add Equations (1), (2), (3), and (4): $(AP + BP) + (CR + DR) = (AS + BQ) + (CQ + DS)$
5. Regroup Terms on Both Sides: $(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)$
6. Notice that:
   • $AP + BP = AB$
   • $CR + DR = CD$
   • $AS + DS = AD$
   • $BQ + CQ = BC$
7. Substituting these segment sums: $\mathbf{AB + CD = AD + BC}$ Hence Proved.

2. Rider 2: Parallelogram Circumscribing a Circle is a Rhombus

Problem: Prove that a parallelogram circumscribing a circle is a rhombus.

                         A ----------------------- B
                        /                         /
                       /      (  CIRCLE  )       /
                      /                         /
                     D ----------------------- C
                         Opposite sides equal (AB = CD, AD = BC)
                         PROVE ALL 4 SIDES EQUAL: AB = BC = CD = DA!

Formal Proof:

1. Given: A parallelogram $ABCD$ circumscribing a circle with center $O$. Since $ABCD$ is a parallelogram: $\mathbf{AB = CD} \quad \text{and} \quad \mathbf{AD = BC} \quad \text{--- (1)}$
2. To Prove: $ABCD$ is a rhombus (i.e., $AB = BC = CD = DA$).
3. Proof: From Rider 1, for any quadrilateral circumscribing a circle: $AB + CD = AD + BC \quad \text{--- (2)}$
4. Substitute $CD = AB$ and $AD = BC$ from Equation (1) into Equation (2): $AB + AB = BC + BC$ $2AB = 2BC$ $\mathbf{AB = BC} \quad \text{--- (3)}$
5. Combining Equations (1) and (3): $AB = BC = CD = DA$
6. Since all four sides of parallelogram $ABCD$ are equal, <u>$ABCD$ is a Rhombus</u>. Hence Proved.

3. Rider 3: Angle Between Tangents and Chord (ngle PTQ = 2ngle OPQ)

Problem: Two tangents $TP$ and $TQ$ are drawn to a circle with center $O$ from an external point $T$. Prove that: $\mathbf{\angle PTQ = 2\angle OPQ}$

                                      P (Point of Contact)
                                     /|
                                    / |
                                   /  | Chord PQ
                                T +   |
                                   \  |
                                    \ |
                                     \|
                                      Q (Point of Contact)
                                   O (Center)

Formal Proof:

1. Given: A circle with center $O$, an external point $T$, and two tangents $TP$ and $TQ$ touching the circle at $P$ and $Q$. Join $OP, OQ,$ and chord $PQ$.
2. To Prove: $\angle PTQ = 2\angle OPQ$.
3. Proof: Let $\mathbf{\angle PTQ = \theta}$.
4. In $\Delta TPQ$:
   • By Theorem 10.2, $TP = TQ$ (tangents from an external point).
   • Since two sides are equal, $\Delta TPQ$ is an isosceles triangle!
   • Therefore, the base angles are equal: $\angle TPQ = \angle TQP$
5. By the Angle Sum Property of $\Delta TPQ$: $\angle PTQ + \angle TPQ + \angle TQP = 180^\circ$ $\theta + 2\angle TPQ = 180^\circ$ $2\angle TPQ = 180^\circ - \theta$ $\angle TPQ = \frac{180^\circ - \theta}{2} = \mathbf{90^\circ - \frac{\theta}{2}} \quad \text{--- (1)}$
6. Now, by Theorem 10.1, the radius through the point of contact is perpendicular to the tangent: $OP \perp TP \implies \mathbf{\angle OPT = 90^\circ}$
7. Notice that $\angle OPT = \angle OPQ + \angle TPQ$: $\angle OPQ = \angle OPT - \angle TPQ$ $\angle OPQ = 90^\circ - \left(90^\circ - \frac{\theta}{2}\right) = 90^\circ - 90^\circ + \frac{\theta}{2} = \mathbf{\frac{\theta}{2}}$
8. Multiply both sides by $2$: $2\angle OPQ = \theta$ Since $\theta = \angle PTQ$: $\mathbf{\angle PTQ = 2\angle OPQ}$ Hence Proved.

4. Rider 4: Supplementary Angles Subtended by Opposite Sides

Problem: Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle (i.e., $\angle AOB + \angle COD = 180^\circ$).

Proof Outline:

1. Connect center $O$ to all four vertices ($A, B, C, D$) and all four points of contact ($P, Q, R, S$).
2. Eight right-angled triangles are formed around center $O$.
3. By RHS congruence, adjacent triangles sharing a common vertex have equal central angles: $\angle 1 = \angle 2, \quad \angle 3 = \angle 4, \quad \angle 5 = \angle 6, \quad \angle 7 = \angle 8$
4. Since the sum of all angles around a point is $360^\circ$: $2(\angle 2 + \angle 3 + \angle 6 + \angle 7) = 360^\circ$ $\mathbf{\angle AOB + \angle COD = 180^\circ}$ Hence Proved.

5. Summary and Examination Tips

Exam Tip: In Rider 1, when writing the 4 tangent pairs ($AP = AS, BP = BQ, CR = CQ, DR = DS$), ensure you align all segments belonging to side $AB$ and $CD$ on the left-hand side! Mixing sides across the equality sign makes regrouping into $(AP + BP)$ impossible.

Common Mistake: In $\angle PTQ = 2\angle OPQ$, confusing radius $OP$ with tangent $TP$. Remember that $\angle OPT = 90^\circ$, not $\angle OPQ$!