Circles Ex 10.2 Class 10 NCERT: Theorems, Proofs & Solutions

Understanding the Core Theorems for Exercise 10.2

Exercise 10.2 relies heavily on the geometric properties of tangents. The most vital tool in your toolkit is Theorem 10.2, which states: The lengths of tangents drawn from an external point to a circle are equal. Let's visualize this: if P is an external point, and PA and PB are tangents touching the circle at A and B, then PA = PB. We prove this by joining the center O to P, A, and B. In right triangles OAP and OBP, we have OA = OB (radii of the same circle), OP = OP (common hypotenuse), and angle OAP = angle OBP = 90 degrees (by Theorem 10.1). By the RHS congruence criterion, triangle OAP is congruent to triangle OBP, which directly implies PA = PB by CPCT (Corresponding Parts of Congruent Triangles). Understanding this congruence is key because it also tells us that angle OPA = angle OPB (tangents are equally inclined to the line joining the point to the center) and angle AOP = angle BOP.

Step-by-Step Approach to Solve Circles Ex 10.2 Problems

1. Identify and Draw the Diagram — Always start by drawing a clean, labeled diagram. Identify the center of the circle, the external points, and points of contact.
2. Apply Theorem 10.2 (Equal Tangents) — Look for external points. Write down equations showing that tangents from the same external point to the circle are equal in length (e.g., AP = AS).
3. Utilize Right Angles (Theorem 10.1) — Locate radii meeting tangents at points of contact. Mark these angles as 90 degrees to utilize the Pythagoras theorem or trigonometric ratios if side lengths are involved.
4. Formulate Algebraic Expressions — For perimeter or quadrilateral proofs, express sides as sums of tangent segments (like AB = AP + PB) and substitute equal terms systematically.

Board Exam Tips & Common Pitfalls

When solving questions from circles ex 10 2 class 10 ncert, keep these critical points in mind:

1. Always Mention Theorems by Statement: Do not just write 'By Theorem 10.2'. Write the full statement: 'Since lengths of tangents drawn from an external point to a circle are equal...' to secure full marks.
2. Concentric Circles Confusions: For concentric circle questions, remember that the radius of the smaller circle is perpendicular to the chord of the larger circle at the point of contact, which also bisects the chord.
3. Quadrilateral Proof Alignment: When proving AB + CD = AD + BC, ensure you list the equal tangent pairs so that LHS terms group together to form complete sides of the quadrilateral.

Practice Questions with Solutions

• Q: Prove that in two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact. A: Step 1: Let O be the common centre of two concentric circles C1 and C2. Let AB be a chord of the larger circle C1 which touches the smaller circle C2 at the point P. Step 2: Join OP. Since AB is a tangent to C2 at P and OP is the radius, by Theorem 10.1, OP is perpendicular to AB. Step 3: Since AB is a chord of the circle C1 and OP is perpendicular to AB, the perpendicular drawn from the centre of a circle to a chord bisects the chord (standard class 9 circle property). Final answer: Therefore, AP = PB, which proves that the chord is bisected at the point of contact.
• Q: Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that angle PTQ = 2 angle OPQ. A: Step 1: Let the given angle PTQ = theta. Since TP = TQ (tangents from external point T), triangle TPQ is an isosceles triangle. Step 2: In triangle TPQ, angle TPQ = angle TQP = (1/2)(180 - theta) = 90 - (1/2)theta. Step 3: By Theorem 10.1, the radius OP is perpendicular to tangent TP. Thus, angle OPT = 90 degrees. Step 4: Now, angle OPQ = angle OPT - angle TPQ = 90 - (90 - (1/2)theta) = (1/2)theta = (1/2)angle PTQ. Final answer: Rearranging the equation, we get angle PTQ = 2 * angle OPQ. Hence proved.
• Q: A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC. A: Step 1: Let the circle touch the sides AB, BC, CD, AD at points P, Q, R, S respectively. Step 2: Since lengths of tangents from an external point are equal, we write: - Tangents from A: AP = AS - Tangents from B: BP = BQ - Tangents from C: CR = CQ - Tangents from D: DR = DS Step 3: Add all four equations: (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ). Step 4: Simplify the grouped terms: AP + BP = AB, CR + DR = CD, AS + DS = AD, and BQ + CQ = BC. Final answer: Substituting these back gives AB + CD = AD + BC. Hence proved.
• Q: Prove that the angle between two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre. A: Step 1: Let P be an external point, and PA and PB be tangents to a circle with centre O touching at A and B. Step 2: By Theorem 10.1, OA is perpendicular to PA and OB is perpendicular to PB. Therefore, angle OAP = 90 degrees and angle OBP = 90 degrees. Step 3: In quadrilateral OAPB, the sum of all interior angles is 360 degrees. So, angle APB + angle OAP + angle AOB + angle OBP = 360 degrees. Step 4: Substitute the 90-degree values: angle APB + 90 + angle AOB + 90 = 360 degrees. Final answer: Simplifying this, we get angle APB + angle AOB = 360 - 180 = 180 degrees. This proves the angles are supplementary.

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