Circles — Andhra Pradesh (SSC) Class 10 Mathematics Solutions (Free)

Theorem: if a line touches a circle at exactly one point (a tangent), the radius drawn to that point of contact is perpendicular to the tangent. Given: circle with centre O, tangent line AB touching the circle at point P, and OP the radius to the point of contact. To prove: OP ⊥ AB Proof (by contradiction): Suppose OP is NOT perpendicular to AB. Then there must be some OTHER point Q on the tangent line such that OQ is perpendicular to AB, and by the shortest-distance property, OQ < OP (the perpendicular distance from a point to a line is always the shortest possible distance). But Q lies on the tangent line, and if OQ < OP (the radius), then Q would lie INSIDE the circle. This is impossible, because a tangent, by definition, touches the circle at only ONE point (P) and does not enter the circle's interior at any other point. This contradiction means our assumption was false — there is no such point Q with a shorter perpendicular distance, so OP itself must already be the perpendicular from O to the tangent line. Hence, OP ⊥ AB — the radius is perpendicular to the tangent at the point of contact.

Given: circle with centre O, and an external point P from which two tangents PA and PB touch the circle at points A and B respectively. To prove: PA = PB Proof: Join OA, OB, and OP. Since PA and PB are tangents, by the previous theorem, OA ⊥ PA and OB ⊥ PB — so ∠OAP = ∠OBP = 90°. Now consider triangles OAP and OBP: - OA = OB (both are radii of the same circle) - OP = OP (common side, shared by both triangles) - ∠OAP = ∠OBP = 90° By the RHS (Right angle–Hypotenuse–Side) congruence criterion, △OAP ≅ △OBP. Since the triangles are congruent, their corresponding sides are equal: PA = PB. Hence, the two tangents drawn from an external point to a circle are always equal in length.

Since a tangent is always perpendicular to the radius at the point of contact, the radius (OA), the tangent (PA), and the line joining the centre to the external point (OP) form a right-angled triangle, right-angled at A. Given: OA (radius) = 9 cm, OP = 15 cm. Using the Pythagorean theorem: OP² = OA² + PA² PA² = OP² − OA² = 15² − 9² = 225 − 81 = 144 PA = √144 = 12 cm By the earlier theorem, both tangents from P are equal in length, so each tangent measures 12 cm.

The perpendicular drawn from a circle's centre to any chord always BISECTS that chord — this is a standard circle theorem. Given: chord length = 16 cm, radius = 10 cm. Step 1: Half of the chord = 16/2 = 8 cm Step 2: The radius, the half-chord, and the perpendicular distance from the centre to the chord form a right-angled triangle, with the radius as the hypotenuse. Using the Pythagorean theorem: (distance)² = radius² − (half-chord)² distance² = 10² − 8² = 100 − 64 = 36 distance = √36 = 6 cm The chord is 6 cm from the centre.

When two circles touch each other EXTERNALLY (each circle lies outside the other, touching at exactly one point), the point of contact lies on the straight line joining the two centres, positioned exactly between them. This means the distance between the centres is simply the SUM of the two radii: Distance = r₁ + r₂ = 7 + 4 = 11 cm The distance between the centres is 11 cm. (Note: if the circles instead touched INTERNALLY — one circle inside the other — the distance between centres would instead be the DIFFERENCE of the radii, not the sum.)

Since the secant passes directly through the centre O, points A and B lie on opposite ends of a diameter, both along the same line as P and O. Given: OP = 20 cm, radius = 7 cm. PA (distance from P to the NEARER intersection point) = OP − radius = 20 − 7 = 13 cm (Point A lies between P and O, at radius-distance from O, so it is closer to P by exactly one radius than O itself.) PB (distance from P to the FARTHER intersection point) = OP + radius = 20 + 7 = 27 cm (Point B lies on the opposite side of O from P, so reaching it means travelling the full distance to the centre AND then one more radius beyond it.) PA = 13 cm, PB = 27 cm.

When a chord subtends a known angle θ at the centre, the chord length can be found using: Chord length = 2r sin(θ/2) where r is the radius. This formula comes directly from splitting the isosceles triangle formed by the two radii and the chord into two equal right-angled triangles using the perpendicular bisector from the centre. Given: r = 8 cm, θ = 90° Chord = 2 × 8 × sin(45°) sin(45°) = √2/2 ≈ 0.707 Chord = 16 × 0.707 ≈ 11.31 cm Exactly expressed: Chord = 16 × (√2/2) = 8√2 cm ≈ 11.31 cm.

This uses the Intersecting Chords Theorem: when two chords of a circle intersect inside the circle, the products of the lengths of their respective segments are equal. AP × PB = CP × PD Given: AP = 5 cm, PB = 8 cm, CP = 4 cm. Find PD. 5 × 8 = 4 × PD 40 = 4 × PD PD = 40/4 = 10 cm The length of PD is 10 cm. (This theorem works because triangles APC and DPB are always similar to each other whenever two chords intersect this way, which leads directly to this product relationship between their segments.)