Circles — Karnataka (SSLC) Class 10 Mathematics Solutions (Free)

Theorem: A tangent to a circle is perpendicular to the radius at the point of tangency. Given: Let O be the center of a circle. Let AB be a tangent to the circle at point P. OP is the radius to point P. Prove: OP ⊥ AB (that is, OP is perpendicular to AB) Proof: Step 1: Assume, for contradiction, that OP is not perpendicular to AB. Suppose OP makes an angle other than 90° with the tangent AB. Step 2: From O, draw a perpendicular to the tangent AB, meeting it at point Q. Since this perpendicular from O to line AB meets at Q, we have OQ ⊥ AB. Step 3: Now, OQ is perpendicular to the tangent, so OQ is less than or equal to OP. In fact, if Q ≠ P, then by the geometry of perpendiculars, OQ < OP. Step 4: But we know OP is a radius to the point of contact P. The definition of a tangent is a line that touches the circle at exactly one point and does not intersect the circle at any other point. Step 5: If OQ < OP and Q is on the tangent line, then Q must be inside the circle (since Q is closer to O than P is). But if Q is on the tangent line and inside the circle, this contradicts the fact that the tangent touches the circle at only one point. Step 6: Therefore, our assumption must be false. We must have Q = P, which means the perpendicular from O to the tangent at the point of contact coincides with OP. Step 7: Conclusion. OP ⊥ AB Hence proved. ✓ Converse: If a line is perpendicular to a radius at the point where the radius meets the circle, then the line is a tangent to the circle.

Theorem: Two tangents drawn to a circle from an external point are equal in length. Given: Let O be the center of the circle. Let P be an external point. PA and PB are two tangents to the circle at points A and B respectively. Prove: PA = PB Proof: Step 1: Since PA and PB are tangents, we use the perpendicularity property. OA ⊥ PA (radius is perpendicular to tangent at point of contact) OB ⊥ PB (radius is perpendicular to tangent at point of contact) Step 2: Consider triangles OAP and OBP. In triangle OAP: ∠OAP = 90° (from Step 1) In triangle OBP: ∠OBP = 90° (from Step 1) Step 3: Identify equal sides. OA = OB (both are radii of the same circle) OP = OP (common side) Step 4: Apply the RHS (Right Angle - Hypotenuse - Side) congruence criterion. In triangles OAP and OBP: ∠OAP = ∠OBP = 90° OP = OP (hypotenuse is common) OA = OB (one side) By RHS congruence: Triangle OAP ≅ Triangle OBP Step 5: From congruent triangles, corresponding parts are equal. PA = PB (corresponding sides of congruent triangles) Hence proved. ✓ Additional Results: 1. The line joining P and O bisects the angle ∠APB. 2. The line PO bisects the chord AB at right angles.

Given: Circle with center O and radius r = 5 cm Point P at distance OP = 13 cm from center Two tangents PA and PB are drawn from P to the circle Find: Length of each tangent (PA and PB) Diagram (text description): A (point of tangency) /| / | / | 5 cm (radius OA) / | / | / O________ / 13 cm / P \_____ B (point of tangency) Solution: Step 1: Since PA is a tangent at point A: OA ⊥ PA Therefore, triangle OAP is a right triangle with right angle at A. Step 2: In right triangle OAP, apply the Pythagorean theorem. OP² = OA² + PA² 13² = 5² + PA² 169 = 25 + PA² PA² = 169 - 25 PA² = 144 PA = 12 cm Step 3: Similarly, for tangent PB at point B: OB ⊥ PB Triangle OBP is right-angled at B. Step 4: In right triangle OBP: OP² = OB² + PB² 13² = 5² + PB² 169 = 25 + PB² PB² = 144 PB = 12 cm Alternatively, by the theorem proved earlier: PA = PB (tangents from an external point are equal) So PB = 12 cm Answer: Length of each tangent = 12 cm Verification: 13² = 5² + 12² 169 = 25 + 144 169 = 169 ✓ (Pythagorean triple: 5-12-13)

Given: Length of chord = 24 cm Radius of circle = 13 cm Find: Distance of chord from the center Diagram (text description): O (center) | | d (perpendicular distance) | ----M---- (M is foot of perpendicular, midpoint of chord) / | \ A | B |--12-|-12--| (half-chord on each side) \__________/ Chord AB = 24 cm Key Property: The perpendicular from the center of a circle to a chord bisects the chord. Step 1: Let O be the center and AB be the chord with length 24 cm. Let M be the foot of the perpendicular from O to AB. M is the midpoint of AB. Step 2: Find AM (half the chord). AM = AB / 2 = 24 / 2 = 12 cm Step 3: In right triangle OMA: ∠OMA = 90° (perpendicular) OA = 13 cm (radius) AM = 12 cm Step 4: Apply the Pythagorean theorem. OA² = OM² + AM² 13² = OM² + 12² 169 = OM² + 144 OM² = 169 - 144 OM² = 25 OM = 5 cm Answer: Distance of chord from the center = 5 cm Verification: 13² = 5² + 12² 169 = 25 + 144 169 = 169 ✓ (Pythagorean triple: 5-12-13)

Given: Circle 1: Center O₁, radius r₁ = 8 cm Circle 2: Center O₂, radius r₂ = 5 cm The circles touch each other externally Find: Distance between centers O₁O₂ Diagram (text description): Circle 1 Point of tangency Circle 2 * * * \ / \ / \ 8 cm / \ 5 cm / \ O₁ O₂ / *-----------•---------•-----------* 8 cm ? 5 cm Definition: Two circles touch externally when they have exactly one point in common, and one circle lies outside the other. Step 1: Understand external tangency. When two circles touch externally: - The point of tangency lies on the line joining the centers - The distance between centers = sum of the radii Step 2: Apply the formula for external tangency. Distance between centers = r₁ + r₂ O₁O₂ = 8 + 5 O₁O₂ = 13 cm Answer: Distance between centers = 13 cm Note: For internal tangency (one circle inside the other, touching at one point): Distance between centers = |r₁ - r₂| = |8 - 5| = 3 cm

Given: Circle with center O and radius r = 10 cm Point P at distance OP = 26 cm from center Secant through P and O intersects circle at A and B A is closer to P than B Find: PA and PB Diagram (text description): A O B •----•---------•----P 10 cm 10 cm 16 cm Secant line through O Solution: Step 1: Understand the configuration. Since the secant passes through the center O, it's actually a diameter extended to point P. Let's place O at the origin with the secant line horizontal. Step 2: Identify positions on the secant. On the line POB (from left to right): - A is on the circle, on the side closer to P - O is the center - B is on the circle, on the opposite side - P is external Actually, let's reconsider. If OP = 26 and the radius is 10: - A is at distance OA = 10 on one side - B is at distance OB = 10 on the opposite side - P is at distance OP = 26 from O Step 3: Since the secant passes through O: PA = OP - OA = 26 - 10 = 16 cm (This is the distance from P to the near intersection point) PB = OP + OB = 26 + 10 = 36 cm (This is the distance from P to the far intersection point) Answer: PA = 16 cm PB = 36 cm Verification using Power of a Point: Power of point P = PA × PB = 16 × 36 = 576 Also, Power of point P = OP² - r² = 26² - 10² = 676 - 100 = 576 ✓

Given: Radius of circle = 6 cm Angle subtended by chord at center = 60° Find: Length of chord Diagram (text description): O (center) /| / | 6 cm (radius) / | 6 cm | 60° angle / | A-----B (chord) Solution: Step 1: Consider the triangle OAB formed by the two radii and the chord. OA = OB = 6 cm (radii) ∠AOB = 60° (angle at center) Step 2: Since OA = OB, triangle OAB is isosceles. Step 3: Find the base angles. In isosceles triangle OAB: ∠OAB = ∠OBA = (180° - 60°) / 2 = 120° / 2 = 60° Step 4: Since all angles are 60°, the triangle is equilateral. Step 5: In an equilateral triangle, all sides are equal. AB = OA = OB = 6 cm Answer: Length of chord = 6 cm Alternative Method using Chord Length Formula: For a chord subtending angle θ at the center of a circle with radius r: Chord length = 2r sin(θ/2) Chord length = 2 × 6 × sin(60°/2) = 12 × sin(30°) = 12 × (1/2) = 6 cm ✓

Given: Two chords AB and CD intersect at point P inside a circle AP = 4 cm, PB = 6 cm, CP = 3 cm Find: PD Diagram (text description): C | / | \ / | \ / | \ A-----P-----B \ | / \ | / \ | / D Theorem (Intersecting Chords): If two chords of a circle intersect at a point P inside the circle, then: AP × PB = CP × PD This is known as the Power of a Point theorem for intersecting chords. Step 1: Apply the intersecting chords theorem. AP × PB = CP × PD 4 × 6 = 3 × PD 24 = 3 × PD Step 2: Solve for PD. PD = 24 / 3 PD = 8 cm Answer: PD = 8 cm Verification: AP × PB = 4 × 6 = 24 CP × PD = 3 × 8 = 24 ✓ Note: This theorem applies to any two chords intersecting inside a circle, regardless of whether they're perpendicular or at any other angle.