Chapter 10: Circles

Solution

A tangent is always perpendicular to the radius at the point of contact. Angle = 90°.

Solution

A point inside the circle cannot have any tangent line to the circle passing through it.

Solution

PA² = OP² − r² = 169 − 25 = 144. PA = 12 cm.

Solution

∠OAP = 90°. OP² = OA² + PA² = 9 + 16 = 25. OP = 5 cm.

Solution

∠AOB + ∠APB = 180° → ∠AOB = 180° − 70° = 110°.

Solution

The single point where a tangent touches the circle is called the point of contact (or point of tangency).

Solution

∠OAP = ∠OBP = 90°. In quadrilateral OAPB: ∠AOB + ∠APB = 360° − 180° = 180°.

∠AOB = 180° − 60° = 120°.

Solution

Semi-perimeter s = (3+4+5)/2 = 6. Area = ½×3×4 = 6 (right triangle).

r = Area/s = 6/6 = 1 cm.

Solution

Let PA, PB be tangents from P, with centre O. OA = OB (radii), OP common, ∠OAP = ∠OBP = 90°. By RHS: △OAP ≅ △OBP → PA = PB.

Solution

Let the circle touch AB at P, BC at Q, CD at R, DA at S.

AP = AS, BP = BQ, CR = CQ, DR = DS (tangent lengths from external points).

AB + CD = AP + PB + CR + RD = AS + BQ + CQ + DS = (AS+DS) + (BQ+CQ) = AD + BC.

Solution

The chord is tangent to the inner circle, so the perpendicular from centre = 3 cm.

Half-chord = √(25−9) = 4 cm. Chord = 8 cm.

Solution

∠OPA = 30° (OP bisects ∠APB). cos30° = PA/OP → √3/2 = 6/OP → OP = 12/√3 = 4√3.

Solution

By alternate segment theorem, ∠QPT = ∠QRP (angle in alternate segment). So ∠PRQ = 60°.

Solution

Let tangent XY touch at P, X'Y' touch at Q. ∠OPA = ∠OCA = 90° (tangent⊥radius). In △OPA and △OCA: OP=OC (radii), OA common, AP=AC (tangent from A). By SSS, △OPA ≅ △OCA → ∠POA = ∠COA.

Similarly ∠QOB = ∠COB. ∠POA + ∠COA + ∠COB + ∠QOB = 180° (POQ is diameter, straight line). 2∠COA + 2∠COB = 180° → ∠AOB = ∠COA + ∠COB = 90°.

Solution

r² = (5√2)² − 5² = 50 − 25 = 25. r = 5 cm.

Solution

∠AOB = 130°. ∠APB = 180° − 130° = 50°.

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{14^2 - 7^2} = 12.12\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{12^2 - 11^2} = 4.8\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{19^2 - 7^2} = 17.66\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{18^2 - 3^2} = 17.75\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{11^2 - 9^2} = 6.32\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{12^2 - 7^2} = 9.75\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{11^2 - 12^2} = ERROR: math domain error\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{20^2 - 4^2} = 19.6\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{14^2 - 4^2} = 13.42\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{6^2 - 5^2} = 3.32\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{17^2 - 6^2} = 15.91\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{23^2 - 7^2} = 21.91\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{16^2 - 12^2} = 10.58\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{18^2 - 5^2} = 17.29\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{8^2 - 3^2} = 7.42\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{19^2 - 12^2} = 14.73\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{9^2 - 8^2} = 4.12\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{19^2 - 8^2} = 17.23\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{6^2 - 10^2} = ERROR: math domain error\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{8^2 - 5^2} = 6.24\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{8^2 - 12^2} = ERROR: math domain error\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{20^2 - 7^2} = 18.73\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{23^2 - 10^2} = 20.71\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{7^2 - 7^2} = 0.0\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{18^2 - 12^2} = 13.42\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{10^2 - 8^2} = 6.0\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{6^2 - 6^2} = 0.0\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{25^2 - 12^2} = 21.93\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{11^2 - 7^2} = 8.49\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{10^2 - 10^2} = 0.0\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{11^2 - 6^2} = 9.22\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{16^2 - 5^2} = 15.2\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{19^2 - 9^2} = 16.73\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{15^2 - 12^2} = 9.0\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{14^2 - 9^2} = 10.72\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{24^2 - 7^2} = 22.96\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{23^2 - 12^2} = 19.62\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{9^2 - 7^2} = 5.66\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{16^2 - 9^2} = 13.23\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{22^2 - 5^2} = 21.42\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{13^2 - 12^2} = 5.0\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{25^2 - 10^2} = 22.91\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{9^2 - 5^2} = 7.48\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{7^2 - 10^2} = ERROR: math domain error\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{21^2 - 7^2} = 19.8\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{13^2 - 8^2} = 10.25\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{22^2 - 6^2} = 21.17\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{16^2 - 3^2} = 15.72\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{20^2 - 3^2} = 19.77\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{20^2 - 10^2} = 17.32\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{21^2 - 10^2} = 18.47\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{6^2 - 7^2} = ERROR: math domain error\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{14^2 - 6^2} = 12.65\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{8^2 - 9^2} = ERROR: math domain error\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{19^2 - 6^2} = 18.03\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{14^2 - 10^2} = 9.8\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{6^2 - 8^2} = ERROR: math domain error\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{24^2 - 12^2} = 20.78\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{13^2 - 5^2} = 12.0\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{12^2 - 3^2} = 11.62\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{24^2 - 8^2} = 22.63\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{18^2 - 11^2} = 14.25\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{18^2 - 6^2} = 16.97\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{15^2 - 7^2} = 13.27\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{15^2 - 11^2} = 10.2\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{16^2 - 4^2} = 15.49\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{15^2 - 5^2} = 14.14\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{24^2 - 5^2} = 23.47\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{22^2 - 4^2} = 21.63\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{24^2 - 4^2} = 23.66\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{25^2 - 4^2} = 24.68\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{17^2 - 3^2} = 16.73\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{22^2 - 9^2} = 20.07\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{17^2 - 8^2} = 15.0\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{24^2 - 9^2} = 22.25\) cm

Solution

Tangent is perpendicular to radius at point of contact.

By Pythagoras: \(L = \sqrt{d^2 - r^2} = \sqrt{20^2 - 12^2} = 16.0\) cm

Solution

BD = 8, DC = 6, r = 4. Let AE = AF = x (tangent lengths, E on AB, F on AC).

AB = x+8, AC = x+6, BC = 14. s = (x+8+x+6+14)/2 = x+14.

Area = r×s = 4(x+14). Also Area by Heron's: s = x+14, s−a = x, s−b = 8, s−c = 6.

Area = √[(x+14)(x)(8)(6)] = √[48x(x+14)].

4(x+14) = √[48x(x+14)] → 16(x+14)² = 48x(x+14) → 16(x+14) = 48x → x+14 = 3x → x = 7.

AB = 15, AC = 13.

Solution

ABCD circumscribes circle. AB+CD = AD+BC (proved earlier). For parallelogram: AB = CD, AD = BC. So 2AB = 2AD → AB = AD. All sides equal → rhombus.

Solution

AB + CD = BC + AD → 6 + 4 = 7 + AD → AD = 3 cm.

Solution

In quadrilateral PAOB: ∠PAO = ∠PBO = 90°. Sum of angles = 360°.

∠APB + ∠AOB + 90° + 90° = 360° → ∠APB + ∠AOB = 180°. Hence supplementary.