Chapter 10Circles

A tangent can be drawn at every single point on a circle, and a circle has infinitely many points.

(i) one (that's the definition of a tangent)
(ii) secant
(iii) two (one on each side, both parallel to a given direction — a third parallel line would have to cross one of them)
(iv) point of contact

Since PQ is tangent at P, OP ⊥ PQ (Theorem 10.1), so ΔOPQ is right-angled at P.
By Pythagoras: PQ² = OQ² − OP² = 12² − 5² = 144 − 25 = 119
PQ = √119

Draw any circle with centre O. Take a line l in any direction outside the circle.
Draw line m through the point on the circle nearest to (or farthest from) O in the direction parallel to l — this line touches the circle at exactly one point, so m is a tangent parallel to l.
Draw line n through the centre O, also parallel to l — since it passes through the centre, it must cut the circle at two points, so n is a secant parallel to l.

The radius is perpendicular to the tangent at the point of contact (Theorem 10.1), so the radius, the tangent, and OQ form a right triangle with OQ as hypotenuse.
r² = OQ² − (tangent)² = 25² − 24² = 625 − 576 = 49 → r = 7

OPTQ is a quadrilateral with ∠OPT = ∠OQT = 90° (tangent ⊥ radius at each point of contact).
The angle sum of a quadrilateral is 360°, so ∠PTQ = 360° − 90° − 90° − 110° = 70°

By the same quadrilateral argument as Q2: ∠AOB = 360° − 90° − 90° − 80° = 100°.
OP bisects ∠AOB (ΔOPA ≡ ΔOPB by Theorem 10.2's proof, so ∠AOP = ∠BOP), so ∠POA = 100°2 = 50°

Let AB be a diameter of a circle with centre O, with tangents drawn at A and at B.
By Theorem 10.1, the tangent at A is perpendicular to OA, and the tangent at B is perpendicular to OB.
Since A, O, B are collinear (AB is a diameter, a straight line through O), OA and OB point along the same line — so both tangents are perpendicular to the same line AB.
Two lines perpendicular to the same line are parallel to each other.

Let XY be the tangent to a circle with centre O, touching it at P. Suppose, for contradiction, that the perpendicular to XY at P does not pass through O — then this perpendicular line and the segment OP would be two different lines both starting at P.
But by Theorem 10.1, OP itself is perpendicular to XY at P. Since there can only be one line perpendicular to XY at the single point P (in a plane, only one perpendicular can be drawn to a line at a given point on it), the perpendicular at P must be the line OP itself.

The radius, tangent, and the line to A form a right triangle (right angle at the point of contact).
r² = OA² − (tangent)² = 5² − 4² = 25 − 16 = 9 → r = 3

Let AB be the chord of the larger circle (radius 5 cm) that touches the smaller circle (radius 3 cm) at P.
Since AB is tangent to the smaller circle at P, OP ⊥ AB (Theorem 10.1), and OP = 3 cm (the smaller radius).
In right ΔOPA, OA = 5 cm (larger radius), OP = 3 cm, so AP² = OA² − OP² = 25 − 9 = 16 → AP = 4 cm.
Since the perpendicular from the centre bisects the chord, AB = 2 × AP = 8 cm.

Let the circle touch sides AB, BC, CD, DA at P, Q, R, S respectively.
By Theorem 10.2, tangents from the same external point are equal in length:
AP = AS (tangents from A), BP = BQ (tangents from B), CR = CQ (tangents from C), DR = DS (tangents from D)
Adding all four: AP + BP + CR + DR = AS + BQ + CQ + DS
(AP+BP) + (CR+DR) = (AS+DS) + (BQ+CQ)
AB + CD = AD + BC

OA bisects ∠PAC (angle between the two tangents from A, i.e. AP along XY and AC), so let ∠OAC = ∠OAP = x.
Similarly, OB bisects ∠QBC, so let ∠OBC = ∠OBQ = y.
Since XY ∥ X′Y′ and AB is a transversal, ∠PAB + ∠QBA = 180° (co-interior angles), i.e. 2x + 2y = 180° → x+y = 90°.
In ΔAOB, the angle sum gives ∠AOB = 180° − (∠OAB + ∠OBA) = 180° − (x+y) = 180° − 90° = 90°

Let PA and PB be tangents from external point P, touching the circle at A and B, with centre O.
∠OAP = ∠OBP = 90° (Theorem 10.1, at each point of contact).
In quadrilateral OAPB, the angle sum is 360°:
∠AOB + ∠OAP + ∠APB + ∠OBP = 360°
∠AOB + 90° + ∠APB + 90° = 360°
∠AOB + ∠APB = 180°

Let ABCD be a parallelogram circumscribing a circle, touching sides AB, BC, CD, DA at P, Q, R, S respectively.
By Theorem 10.2 (as in Q8): AB + CD = AD + BC.
Since ABCD is a parallelogram, AB = CD and AD = BC (opposite sides of a parallelogram are equal).
Substituting: AB + AB = AD + AD → 2AB = 2AD → AB = AD.
So all four sides are equal: AB = CD = AD = BC.

Let the circle touch AB at F and AC at E. By Theorem 10.2: AF = AE = x (say), BF = BD = 8 cm, CE = CD = 6 cm.
So AB = x+8, AC = x+6, BC = 14, and the semi-perimeter s = AB+AC+BC)/2 = x+14.
Using Heron's formula, Area² = s(s−a)(s−b)(s−c) = (x+14)(x)(8)(6) = 48x(x+14).
Also, Area = r × s = 4(x+14) (Area of a triangle = inradius × semi-perimeter).
Squaring: 16(x+14)² = 48x(x+14) → 16(x+14) = 48x → 16x+224 = 48x → 32x = 224 → x = 7

Let ABCD circumscribe a circle with centre O, touching AB, BC, CD, DA at P, Q, R, S.
By the same congruent-triangle argument used in Theorem 10.2, OP bisects ∠APB... more directly: joining O to each vertex, ΔAOP ≡ ΔAOS, ΔBOP ≡ ΔBOQ, ΔCOQ ≡ ΔCOR, ΔDOR ≡ ΔDOS (each pair shares equal tangent-length sides and the common radius, by SSS).
Let ∠AOP = ∠AOS = p, ∠BOP = ∠BOQ = q, ∠COQ = ∠COR = r₁, ∠DOR = ∠DOS = s.
Around O, the full angle is 360°: 2p + 2q + 2r₁ + 2s = 360° → p+q+r₁+s = 180°.
∠AOB = p+q (the angle subtended by side AB) and ∠COD = r₁+s (subtended by side CD).
So ∠AOB + ∠COD = p+q+r₁+s = 180°.